GSEB Class 12 Maths Solutions Chapter 7 સંકલન Exercise 7.1

નીચે આપેલાં વિધેયોના પ્રતિવિકલિત (અનિયત સંકલિત) નિરીક્ષણની રીતે શોધો :

Question 1. \( \sin 2x \)
Answer: ધારો કે \( f(x) = \sin 2x \).
\( f(x) \) નું પ્રતિવિકલિત \( -\frac{1}{2}\cos 2x \) મળે છે.
ચેઇન રૂલનો ઉપયોગ કરીને, આપણે ચકાસી શકીએ છીએ:
\( \frac{d}{dx} \left( -\frac{1}{2}\cos 2x \right) = -\frac{1}{2} (-\sin 2x) \cdot 2 = \sin 2x \)
તેથી, \( F(x) = -\frac{1}{2}\cos 2x + c \), જ્યાં \( c \) એ એક અચળ સંખ્યા છે.
આમ, \( f(x) = \sin 2x \) નું પ્રતિવિકલિત \( F(x) = -\frac{1}{2}\cos 2x + c \) થાય છે.
\( \int \sin 2x \, dx = -\frac{1}{2}\cos 2x + c \)
In simple words: First, we assume the given function is \( f(x) \). Then we find the antiderivative for \( f(x) \). We verify this by differentiating the antiderivative, which brings us back to the original function. Finally, we add a constant \( c \) to get the complete antiderivative.

Exam Tip: Remember to always include the constant of integration, \( c \), when finding an indefinite integral. Also, mentally or explicitly differentiate your answer to check its correctness.

Question 2. \( \cos 3x \)
Answer: ધારો કે \( f(x) = \cos 3x \).
\( f(x) \) નું પ્રતિવિકલિત \( \frac{\sin 3x}{3} \) પ્રાપ્ત થાય છે.
આપણે ચેઇન રૂલનો ઉપયોગ કરીને ચકાસી શકીએ છીએ કે:
\( \frac{d}{dx} \left( \frac{\sin 3x}{3} \right) = \frac{1}{3} \cdot (\cos 3x) \cdot 3 = \cos 3x \)
તેથી, \( F(x) = \frac{\sin 3x}{3} + c \), જ્યાં \( c \) એ અચળ પદ છે.
આમ, \( \int \cos 3x \, dx = \frac{\sin 3x}{3} + c \).
In simple words: We take the function as \( f(x) \). Then we find the antiderivative, which is \( \frac{\sin 3x}{3} \). When we check this by taking its derivative, we get back the original function, confirming our answer. Don't forget to add \( c \).

Exam Tip: For functions like \( \cos(ax+b) \) or \( \sin(ax+b) \), remember to divide by the coefficient of \( x \) (i.e., \( a \)) when integrating.

Question 3. \( e^{2x} \)
Answer: ધારો કે \( f(x) = e^{2x} \).
\( f(x) \) નું પ્રતિવિકલિત \( \frac{e^{2x}}{2} \) મળે છે.
ચકાસણી કરવા માટે, આપણે તેનું વિકલન કરીએ છીએ:
\( \frac{d}{dx} \left( \frac{e^{2x}}{2} \right) = \frac{1}{2} \cdot e^{2x} \cdot 2 = e^{2x} \)
તેથી, \( F(x) = \frac{e^{2x}}{2} + c \), જ્યાં \( c \) એક અચળ પદ છે.
આમ, \( \int e^{2x} \, dx = \frac{e^{2x}}{2} + c \).
In simple words: We consider the function \( f(x) \) as \( e^{2x} \). Its antiderivative is found to be \( \frac{e^{2x}}{2} \). We can check this by differentiating it, which gives us back the original function. Remember to add the constant \( c \).

Exam Tip: The integral of \( e^{ax+b} \) is \( \frac{e^{ax+b}}{a} + c \). This is a very common formula, so make sure you know it well.

Question 4. \( (ax + b)^2 \)
Answer: ધારો કે \( f(x) = (ax + b)^2 \).
\( f(x) \) નું પ્રતિવિકલિત \( \frac{(ax+b)^3}{3a} \) મળે છે.
આપણે ચેઇન રૂલનો ઉપયોગ કરીને ચકાસી શકીએ છીએ:
\( \frac{d}{dx} \left( \frac{(ax + b)^3}{3a} \right) = \frac{1}{3a} \cdot 3(ax + b)^2 \cdot a = (ax + b)^2 \)
તેથી, \( F(x) = \frac{(ax+b)^3}{3a} + c \), જ્યાં \( c \) એક અચળ પદ છે.
આમ, \( \int (ax + b)^2 \, dx = \frac{(ax+b)^3}{3a} + c \).
In simple words: We identify the function as \( (ax+b)^2 \). Its antiderivative is calculated as \( \frac{(ax+b)^3}{3a} \). We confirm this by differentiating it, which gives us the original function back. Always include the integration constant \( c \).

Exam Tip: For functions of the form \( (ax+b)^n \), the integral is \( \frac{(ax+b)^{n+1}}{a(n+1)} + c \). Remember to divide by both \( a \) and \( (n+1) \).

Question 5. \( \sin 2x - 4e^{3x} \)
Answer: ધારો કે \( f(x) = \sin 2x - 4e^{3x} \).
\( f(x) \) નું પ્રતિવિકલિત \( -\frac{\cos 2x}{2} - \frac{4}{3} e^{3x} \) મળે છે.
ચકાસણી કરવા માટે, આપણે તેનું વિકલન કરીએ છીએ:
\( \frac{d}{dx} \left( -\frac{\cos 2x}{2} - \frac{4}{3} e^{3x} \right) \)
\( = - \frac{1}{2} \frac{d}{dx} (\cos 2x) - \frac{4}{3} \frac{d}{dx} (e^{3x}) \)
\( = - \frac{1}{2} (-\sin 2x) \cdot 2 - \frac{4}{3} (e^{3x}) \cdot 3 \)
\( = \sin 2x - 4e^{3x} \)
તેથી, \( F(x) = -\frac{\cos 2x}{2} - \frac{4}{3} e^{3x} + c \), જ્યાં \( c \) એક અચળ સંખ્યા છે.
આમ, \( \int (\sin 2x - 4e^{3x}) \, dx = -\frac{\cos 2x}{2} - \frac{4}{3} e^{3x} + c \).
In simple words: We assume the given expression as \( f(x) \). We find its antiderivative by integrating each term separately. To ensure it's correct, we differentiate the result and confirm it matches the original function. We remember to include the constant \( c \).

Exam Tip: Remember that integration is linear, so you can integrate each term of a sum or difference separately. Apply the appropriate integration rules for each part.

નીચેના સંકલિતો શોધો (પ્રશ્નો 6 થી 20)

Question 6. \( \int(4e^{3x} + 1) \, dx \)
Answer: આપણે ગણતરી કરવા માટે આપેલ સંકલન \( I \) તરીકે લઈએ છીએ:
\( I = \int (4e^{3x} + 1) \, dx \)
સંકલનને અલગ-અલગ પદોમાં વિભાજીત કરતા:
\( = \int 4e^{3x} \, dx + \int 1 \, dx \)
નિયમો લાગુ પાડતા:
\( = 4 \int e^{3x} \, dx + \int x^0 \, dx \)
સંકલન કરીને:
\( = 4\frac{e^{3x}}{3} + \frac{x^{0+1}}{0+1} + c \)
(કારણ કે \( \int e^{ax} \, dx = \frac{e^{ax}}{a} \) અને \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \))
અંતિમ જવાબ:
\( = \frac{4}{3}e^{3x} + x + c \)
In simple words: To integrate this expression, we first break it into two simpler parts. Then, we apply the standard integration rules for \( e^{ax} \) and for a constant term (which is \( x^0 \)). After integrating each part, we combine them and add the constant of integration \( c \).

Exam Tip: Remember to treat constants correctly: a constant multiplying a function can be pulled outside the integral, and the integral of a constant is \( kx+c \).

Question 7. \( \int x^2 \left(1 - \frac{1}{x^2}\right) \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int x^2 \left(1 - \frac{1}{x^2}\right) \, dx \)
પ્રથમ, આપણે \( x^2 \) ને કૌંસમાં ગુણીએ છીએ:
\( I = \int \left( x^2 \cdot 1 - x^2 \cdot \frac{1}{x^2} \right) \, dx \)
\( I = \int (x^2 - 1) \, dx \)
હવે, આપણે દરેક પદનું અલગ-અલગ સંકલન કરીએ છીએ:
\( I = \int x^2 \, dx - \int 1 \, dx \)
ઘાતાંક નિયમ અને અચળના સંકલનનો ઉપયોગ કરીને:
\( I = \frac{x^{2+1}}{2+1} - x + c \)
\( I = \frac{x^3}{3} - x + c \)
In simple words: First, we multiply \( x^2 \) inside the bracket to simplify the expression. This gives us \( x^2 - 1 \). Then, we integrate each part separately. The integral of \( x^2 \) is \( \frac{x^3}{3} \), and the integral of \( -1 \) is \( -x \). Finally, we add the constant of integration \( c \).

Exam Tip: Always simplify the integrand algebraically before integrating, if possible. This can turn a complex-looking integral into simpler power rule applications.

Question 8. \( \int x\sqrt{1+2x^2} \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int x\sqrt{1+2x^2} \, dx \)
આપણે આદેશ લઈએ કે \( 1 + 2x^2 = t^2 \). આનું વિકલન કરતા, આપણને મળે છે:
\( 4x \, dx = 2t \, dt \)
તેથી, \( x \, dx = \frac{t}{2} \, dt \).
અને \( t = \sqrt{1+2x^2} \).
મૂળ સંકલનમાં આ કિંમતો મૂકતા:
\( I = \int \sqrt{t^2} \cdot \frac{t}{2} \, dt \)
\( I = \int t \cdot \frac{t}{2} \, dt \)
તે સરળ બનીને આમ લખી શકાય:
\( I = \frac{1}{2} \int t^2 \, dt \)
સંકલન કરીને:
\( I = \frac{1}{2} \cdot \frac{t^3}{3} + c \)
આપણને મળે છે:
\( I = \frac{1}{6} t^3 + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = \frac{1}{6} (1 + 2x^2)^{\frac{3}{2}} + c \)
In simple words: We solve this integral using substitution. We let \( 1+2x^2 \) be \( t^2 \) to simplify the square root. Differentiating this substitution gives us a way to replace \( x \, dx \) with \( t \, dt \). After simplifying and integrating with respect to \( t \), we substitute the value of \( t \) back in terms of \( x \) to get the final answer.

Exam Tip: When using substitution, choose a part of the integrand (often inside a root, power, or denominator) to simplify the expression. Remember to change \( dx \) to \( dt \) (or \( du \)) correctly and substitute back at the end.

Question 9. \( \int (4x + 2)\sqrt{x^2+x+1} \, dx \)
Answer: આપણે આ સંકલન \( I \) ની ગણતરી કરીએ છીએ:
\( I = \int (4x + 2)\sqrt{x^2+x+1} \, dx \)
પ્રથમ, આપણે પદ \( (4x+2) \) માંથી 2 ને સામાન્ય કાઢીએ છીએ:
\( I = \int 2(2x + 1) \sqrt{x^2+x+1} \, dx \)
આપણે \( x^2+x+1 = t^2 \) નો આદેશ લઈએ છીએ. આના વિકલનથી \( (2x+1)dx = 2t \, dt \) મળે છે. આમ, \( t = \sqrt{x^2+x+1} \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int 2 \cdot t \cdot 2t \, dt \)
સરળ કરતા:
\( I = \int 4t^2 \, dt \)
સંકલન કરતા, આપણને મળે છે:
\( I = \frac{4}{3}t^3 + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = \frac{4}{3}(x^2 + x + 1)^{\frac{3}{2}} + c \).
In simple words: We simplify the integral by factoring out 2 from \( (4x+2) \). Then, we use a substitution: let \( x^2+x+1 \) be \( t^2 \). This substitution helps us change \( (2x+1)dx \) to \( 2t \, dt \). After simplifying and integrating with respect to \( t \), we replace \( t \) with its original expression in terms of \( x \) to get the solution.

Exam Tip: Look for a function whose derivative (or a multiple of it) is also present in the integrand. This is a key indicator for using substitution, often setting the inner function or the entire term under a root as \( t \) or \( t^2 \).

Question 10. \( \int \frac{1}{x-\sqrt{x}} \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \frac{1}{x-\sqrt{x}} \, dx \)
છેદમાંથી \( \sqrt{x} \) સામાન્ય કાઢતા, આપણને મળે છે:
\( I = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} \, dx \)
આપણે \( \sqrt{x}-1 = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, આપણને \( \frac{1}{2\sqrt{x}} \, dx = dt \) મળે છે.
તેથી, \( \frac{1}{\sqrt{x}} \, dx = 2dt \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \frac{2dt}{t} \)
સંકલન કરતા, આપણને મળે છે:
\( I = 2 \log |t| + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, જવાબ મળે છે:
\( I = 2 \log |\sqrt{x}-1| + c \).
(જ્યાં \( t = \sqrt{x}-1 \))
In simple words: We first simplify the denominator by factoring out \( \sqrt{x} \). Then, we use substitution by setting \( \sqrt{x}-1 \) equal to \( t \). We differentiate this to replace \( \frac{1}{\sqrt{x}} \, dx \) with \( 2dt \). After integration, we substitute \( t \) back with \( \sqrt{x}-1 \) to get the final answer.

Exam Tip: When dealing with \( \sqrt{x} \) and \( x \) in the denominator, try factoring \( \sqrt{x} \) to create a suitable substitution. The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \), which is useful for substitution.

Question 11. \( \int \frac{x}{\sqrt{x+4}} \, dx, x > -4 \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \frac{x}{\sqrt{x+4}} \, dx \)
આપણે \( x+4 = t \) નો આદેશ લઈએ છીએ. આનાથી \( dx = dt \) મળે છે.
અને \( x \) ને \( t \) ના સ્વરૂપમાં દર્શાવતા: \( x = t-4 \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \frac{t-4}{\sqrt{t}} \, dt \)
પદોને અલગ કરતા:
\( I = \int \left(\frac{t}{\sqrt{t}} - \frac{4}{\sqrt{t}}\right) \, dt \)
ઘાતાંકના રૂપમાં લખતા:
\( I = \int (t^{\frac{1}{2}} - 4t^{-\frac{1}{2}}) \, dt \)
સંકલન કરીને:
\( I = \frac{t^{\frac{3}{2}}}{\frac{3}{2}} - 4\frac{t^{\frac{1}{2}}}{\frac{1}{2}} + c \)
સરળ કરતા:
\( I = \frac{2}{3}t^{\frac{3}{2}} - 8t^{\frac{1}{2}} + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = \frac{2}{3}(x+4)^{\frac{3}{2}} - 8(x+4)^{\frac{1}{2}} + c \).
(જ્યાં \( t = x+4 \))
In simple words: We use substitution to solve this integral. We let \( x+4 \) be \( t \), which means \( x \) becomes \( t-4 \) and \( dx \) becomes \( dt \). We then substitute these into the integral and simplify the expression into powers of \( t \). After integrating each term, we substitute \( t \) back with \( x+4 \) to get the final solution.

Exam Tip: When the numerator contains \( x \) and the denominator has \( \sqrt{x+a} \), substitute \( u = x+a \). This allows you to express \( x \) as \( u-a \) and simplify the integrand for easier power rule application.

Question 12. \( \int (x^3 - 1)^{\frac{1}{3}} x^5 \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int (x^3 - 1)^{\frac{1}{3}} x^5 \, dx \)
આપણે \( x^3-1 = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( 3x^2 \, dx = dt \) મળે છે.
તેથી, \( x^2 \, dx = \frac{dt}{3} \).
અને \( x^3 \) ને \( t \) ના સ્વરૂપમાં દર્શાવતા: \( x^3 = t+1 \).
સંકલનને ફરીથી ગોઠવતા: \( I = \int (x^3 - 1)^{\frac{1}{3}} \cdot x^3 \cdot x^2 \, dx \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int t^{\frac{1}{3}} (t+1) \frac{dt}{3} \)
\( \frac{1}{3} \) ને બહાર કાઢીને અને પદાવલિને વિસ્તૃત કરતા:
\( I = \frac{1}{3} \int (t^{\frac{4}{3}} + t^{\frac{1}{3}}) \, dt \)
સંકલન કરીને:
\( I = \frac{1}{3} \left( \frac{t^{\frac{7}{3}}}{\frac{7}{3}} + \frac{t^{\frac{4}{3}}}{\frac{4}{3}} \right) + c \)
વધુ સરળ કરતા:
\( I = \frac{1}{7}t^{\frac{7}{3}} + \frac{1}{4}t^{\frac{4}{3}} + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ મળે છે:
\( I = \frac{1}{7}(x^3 - 1)^{\frac{7}{3}} + \frac{1}{4}(x^3 - 1)^{\frac{4}{3}} + c \).
(જ્યાં \( t = x^3-1 \))
In simple words: This integral uses substitution and algebraic manipulation. We set \( x^3-1 \) as \( t \) to simplify the cubic root. Then, we express \( x^3 \) as \( t+1 \) and \( x^2 \, dx \) as \( \frac{dt}{3} \). After substituting, we expand the terms, integrate using the power rule, and finally replace \( t \) with \( x^3-1 \) to arrive at the solution.

Exam Tip: When \( x^n \) is multiplied by a function of \( x^{n+1} \), rewrite \( x^n \) by breaking down the higher power of \( x \) (e.g., \( x^5 = x^3 \cdot x^2 \)) to match the derivative of the substitution.

Question 13. \( \int \frac{x^2}{\left(2+3x^3\right)^3} \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \frac{x^2}{\left(2+3x^3\right)^3} \, dx \)
આપણે \( 2+3x^3 = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( 9x^2 \, dx = dt \) મળે છે.
તેથી, \( x^2 \, dx = \frac{dt}{9} \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \frac{dt}{9t^3} \)
\( \frac{1}{9} \) ને બહાર કાઢીને ઘાતાંકના રૂપમાં લખતા:
\( I = \frac{1}{9} \int t^{-3} \, dt \)
સંકલન કરીને:
\( I = \frac{1}{9} \cdot \frac{t^{-2}}{-2} + c \)
સરળ કરતા:
\( I = - \frac{1}{18} t^{-2} + c \)
ઘાતાંકને ધન બનાવીને:
\( I = - \frac{1}{18t^2} + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = - \frac{1}{18(2+3x^3)^2} + c \).
(જ્યાં \( t = 2+3x^3 \))
In simple words: To solve this integral, we use the substitution method. We let \( 2+3x^3 \) be \( t \). Differentiating this gives us \( 9x^2 \, dx = dt \), which helps us replace \( x^2 \, dx \) with \( \frac{dt}{9} \). The integral then simplifies to a power of \( t \). After integrating and simplifying, we substitute \( t \) back with its expression in terms of \( x \) to get the final answer.

Exam Tip: When you see a function and its derivative (or a multiple of it) in the integrand, consider using substitution. For terms like \( (ax+b)^n \), let \( u = ax+b \).

Question 14. \( \int \frac{1}{x(\log x)^m} \, dx, x > 0, m \ne 1 \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \frac{1}{x(\log x)^m} \, dx \)
આપણે \( \log x = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( \frac{1}{x} \, dx = dt \) મળે છે.
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \frac{1}{t^m} \, dt \)
ઘાતાંકના રૂપમાં લખતા:
\( I = \int t^{-m} \, dt \)
સંકલન કરીને:
\( I = \frac{t^{-m+1}}{-m+1} + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = \frac{(\log x)^{-m+1}}{-m+1} + c \).
(જ્યાં \( t = \log x \))
In simple words: We solve this integral using substitution. We let \( \log x \) be \( t \). The derivative of \( \log x \) is \( \frac{1}{x} \), so \( \frac{1}{x} \, dx \) becomes \( dt \). The integral simplifies to \( t^{-m} \), which we integrate using the power rule. Finally, we substitute \( t \) back with \( \log x \) to get the solution.

Exam Tip: When \( \log x \) is present in the integrand, and \( \frac{1}{x} \) is also there, substituting \( t = \log x \) is often effective because \( dt = \frac{1}{x} \, dx \).

Question 15. \( \int \frac{x}{9-4x^2} \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \frac{x}{9-4x^2} \, dx \)
આપણે \( 9-4x^2 = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( -8x \, dx = dt \) મળે છે.
તેથી, \( x \, dx = - \frac{dt}{8} \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \frac{-dt/8}{t} \)
\( - \frac{1}{8} \) ને બહાર કાઢીને:
\( I = - \frac{1}{8} \int \frac{1}{t} \, dt \)
સંકલન કરીને:
\( I = - \frac{1}{8} \log |t| + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = - \frac{1}{8} \log |9-4x^2| + c \).
(જ્યાં \( t = 9-4x^2 \))
In simple words: We solve this integral using the substitution method. We let the denominator \( 9-4x^2 \) be \( t \). Differentiating this gives us \( -8x \, dx = dt \). We then substitute \( x \, dx \) with \( -\frac{dt}{8} \). The integral simplifies to \( -\frac{1}{8} \int \frac{1}{t} \, dt \), which we integrate to \( -\frac{1}{8} \log|t| \). Finally, we replace \( t \) with \( 9-4x^2 \) to get the solution.

Exam Tip: When the numerator is a multiple of the derivative of the denominator, a logarithmic integral often results. Look for expressions like \( \frac{f'(x)}{f(x)} \), which integrates to \( \log|f(x)| + c \).

Question 16. \( \int e^{2x+3} \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int e^{2x+3} \, dx \)
આપણે \( 2x+3 = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( 2dx = dt \) મળે છે.
તેથી, \( dx = \frac{dt}{2} \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int e^t \frac{dt}{2} \)
\( \frac{1}{2} \) ને બહાર કાઢીને:
\( I = \frac{1}{2} \int e^t \, dt \)
સંકલન કરીને:
\( I = \frac{1}{2} e^t + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = \frac{1}{2} e^{2x+3} + c \).
(જ્યાં \( t = 2x+3 \))
In simple words: To integrate \( e^{2x+3} \), we use substitution by letting \( 2x+3 \) be \( t \). Differentiating this gives us \( 2dx = dt \), so \( dx = \frac{dt}{2} \). The integral becomes \( \frac{1}{2} \int e^t \, dt \). Integrating \( e^t \) gives \( e^t \). Finally, we replace \( t \) with \( 2x+3 \) to get the final answer.

Exam Tip: For integrals involving \( e^{ax+b} \), the simplest substitution is \( u = ax+b \). Remember that the derivative of \( ax+b \) is \( a \), which means you will divide by \( a \) in the final integral.

Question 17. \( \int \frac{x}{e^{x^2}} \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \frac{x}{e^{x^2}} \, dx \)
આપણે \( x^2 = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( 2x \, dx = dt \) મળે છે.
તેથી, \( x \, dx = \frac{dt}{2} \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \frac{dt}{2e^t} \)
\( \frac{1}{2} \) ને બહાર કાઢીને ઘાતાંકના રૂપમાં લખતા:
\( I = \frac{1}{2} \int e^{-t} \, dt \)
સંકલન કરીને:
\( I = \frac{1}{2} \frac{e^{-t}}{-1} + c \)
સરળ કરતા:
\( I = - \frac{1}{2} e^{-t} + c \)
ઘાતાંકને ધન બનાવીને:
\( I = - \frac{1}{2e^t} + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = - \frac{1}{2e^{x^2}} + c \).
(જ્યાં \( t = x^2 \))
In simple words: To integrate this, we use substitution by setting \( x^2 \) equal to \( t \). Differentiating this gives us \( 2x \, dx = dt \), so \( x \, dx \) becomes \( \frac{dt}{2} \). The integral simplifies to \( \frac{1}{2} \int e^{-t} \, dt \). After integrating, we replace \( t \) with \( x^2 \) to obtain the final answer.

Exam Tip: When \( e \) is raised to a power that is a function of \( x \) (e.g., \( x^2 \)), and the derivative of that power (e.g., \( x \)) is also present, substitution is usually the method to choose.

Question 18. \( \int \frac{e^{\tan^{-1} x}}{1+x^2} \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \frac{e^{\tan^{-1} x}}{1+x^2} \, dx \)
આપણે \( \tan^{-1} x = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( \frac{1}{1+x^2} \, dx = dt \) મળે છે.
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int e^t \, dt \)
સંકલન કરીને:
\( I = e^t + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = e^{\tan^{-1} x} + c \).
(જ્યાં \( t = \tan^{-1} x \))
In simple words: We solve this integral using substitution by letting \( \tan^{-1} x \) be \( t \). The derivative of \( \tan^{-1} x \) is \( \frac{1}{1+x^2} \), so \( \frac{1}{1+x^2} \, dx \) becomes \( dt \). The integral simplifies to \( \int e^t \, dt \), which is \( e^t \). Finally, we substitute \( t \) back with \( \tan^{-1} x \) to get the solution.

Exam Tip: Recognize derivatives of inverse trigonometric functions. The presence of \( \frac{1}{1+x^2} \) often suggests a substitution involving \( \tan^{-1} x \).

Question 19. \( \int \frac{e^{2x}-1}{e^{2x}+1} \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \frac{e^{2x}-1}{e^{2x}+1} \, dx \)
આપણે અંશ અને છેદને \( e^x \) વડે ભાગીને સરળ કરીએ છીએ:
\( I = \int \frac{e^x(e^x - e^{-x})}{e^x(e^x + e^{-x})} \, dx \)
જે આ પ્રકારે બને છે:
\( I = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} \, dx \)
આપણે \( e^x + e^{-x} = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( (e^x - e^{-x}) \, dx = dt \) મળે છે.
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \frac{dt}{t} \)
સંકલન કરીને:
\( I = \log |t| + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = \log |e^x + e^{-x}| + c \).
(જ્યાં \( t = e^x + e^{-x} \))
In simple words: First, we divide both the numerator and denominator by \( e^x \) to simplify the expression. Then, we use substitution, setting the denominator \( e^x + e^{-x} \) as \( t \). The derivative of this expression \( (e^x - e^{-x}) \, dx \) becomes \( dt \). The integral simplifies to \( \int \frac{dt}{t} \), which integrates to \( \log|t| \). Finally, we replace \( t \) with \( e^x + e^{-x} \) for the final solution.

Exam Tip: For rational functions involving exponential terms, consider dividing the numerator and denominator by \( e^x \) (or a similar term) to create a form suitable for \( u \)-substitution, often leading to a logarithmic integral.

Question 20. \( \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}} \, dx \)
આપણે \( e^{2x} + e^{-2x} = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( (2e^{2x} - 2e^{-2x}) \, dx = dt \) મળે છે.
આને સરળ કરતા:
\( 2(e^{2x} - e^{-2x}) \, dx = dt \)
તેથી, \( (e^{2x} - e^{-2x}) \, dx = \frac{dt}{2} \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \frac{dt/2}{t} \)
\( \frac{1}{2} \) ને બહાર કાઢીને:
\( I = \frac{1}{2} \int \frac{1}{t} \, dt \)
સંકલન કરીને:
\( I = \frac{1}{2} \log |t| + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = \frac{1}{2} \log |e^{2x} + e^{-2x}| + c \).
(જ્યાં \( t = e^{2x} + e^{-2x} \))
In simple words: This integral uses substitution. We set the denominator \( e^{2x} + e^{-2x} \) as \( t \). Differentiating this expression gives us \( 2(e^{2x} - e^{-2x}) \, dx = dt \), which helps us to substitute the numerator and \( dx \) with \( \frac{dt}{2} \). The integral becomes \( \frac{1}{2} \int \frac{1}{t} \, dt \), which integrates to \( \frac{1}{2} \log|t| \). Finally, we replace \( t \) with its original expression in terms of \( x \) to get the solution.

Exam Tip: Pay attention to constants when differentiating the substitution. If \( u = f(ax+b) \), then \( du = a f'(ax+b) \, dx \). Make sure to adjust the \( dt \) term correctly.

Question 21. \( \int \tan^2(2x - 3) \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \tan^2(2x-3) \, dx \)
આપણે \( 2x-3 = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( 2dx = dt \) મળે છે.
તેથી, \( dx = \frac{dt}{2} \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \tan^2 t \frac{dt}{2} \)
ત્રિકોણમિતિના નિત્યસમ \( \tan^2 t = \sec^2 t - 1 \) નો ઉપયોગ કરતા:
\( I = \frac{1}{2} \int (\sec^2 t - 1) \, dt \)
સંકલનને અલગ-અલગ પદોમાં વિભાજીત કરતા:
\( I = \frac{1}{2} \left( \int \sec^2 t \, dt - \int 1 \, dt \right) \)
સંકલન કરીને:
\( I = \frac{1}{2} (\tan t - t) + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા:
\( I = \frac{1}{2} \tan(2x-3) - \frac{1}{2}(2x-3) + c \)
પદોને સરળ કરતા, આપણને મળે છે:
\( I = \frac{1}{2} \tan(2x-3) - x + \frac{3}{2} + c \)
અચળ પદોને જોડીને, અંતિમ જવાબ આવે છે:
\( I = \frac{1}{2} \tan(2x-3) - x + C \).
In simple words: To integrate \( \tan^2(2x-3) \), we first use substitution by letting \( 2x-3 \) be \( t \). Then, we apply the trigonometric identity \( \tan^2 t = \sec^2 t - 1 \). We integrate \( \sec^2 t \) to \( \tan t \) and \( 1 \) to \( t \). Finally, we substitute \( t \) back with \( 2x-3 \) and combine the constants to get the final answer.

Exam Tip: Remember the fundamental trigonometric identities, especially for \( \tan^2 x \) and \( \cot^2 x \), as they are essential for integrating these functions by converting them into terms whose integrals are known (like \( \sec^2 x \) or \( \csc^2 x \)).

Question 22. \( \int \sec^2(7-4x) \, dx \)
Answer: આપણે આપેલ સંકલનની ગણતરી કરીએ છીએ:
\( I = \int \sec^2(7-4x) \, dx \)
આપણે \( 7-4x = t \) નો આદેશ લઈએ છીએ. આનું વિકલન કરતા, \( -4dx = dt \) મળે છે.
તેથી, \( dx = - \frac{dt}{4} \).
આદેશ મૂકતા, સંકલન આ પ્રકારે બને છે:
\( I = \int \sec^2 t \left(-\frac{dt}{4}\right) \)
\( - \frac{1}{4} \) ને બહાર કાઢીને:
\( I = - \frac{1}{4} \int \sec^2 t \, dt \)
સંકલન કરીને:
\( I = - \frac{1}{4} \tan t + c \)
અંતે, \( t \) ની કિંમત પાછી મૂકતા, અંતિમ જવાબ આવે છે:
\( I = - \frac{1}{4} \tan(7-4x) + c \).
(જ્યાં \( t = 7-4x \))
In simple words: To integrate \( \sec^2(7-4x) \), we use substitution by setting \( 7-4x \) equal to \( t \). Differentiating this gives us \( -4dx = dt \), so \( dx \) becomes \( -\frac{dt}{4} \). The integral simplifies to \( -\frac{1}{4} \int \sec^2 t \, dt \). We integrate \( \sec^2 t \) to \( \tan t \). Finally, we replace \( t \) with \( 7-4x \) to obtain the final answer.

Exam Tip: The integral of \( \sec^2(ax+b) \) is \( \frac{1}{a} \tan(ax+b) + c \). Remember this direct formula for quick solutions, but also understand the substitution process behind it.

Question 23. \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\)
Answer: The initial integral is given as I = \(\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} dx\). We assume that \( \sin^{-1} x \) equals \( t \). This implies that \( \frac{1}{\sqrt{1-x^2}} dx \) becomes \( dt \). The integral then transforms into I = \(\int t dt\). Integrating this expression yields \( \frac{t^2}{2} + c \). Substituting \( t \) back with \( \sin^{-1} x \) gives the final result. The final result for the integral is \( \frac{(\sin^{-1}x)^2}{2} + c \). (since \( t = \sin^{-1}x \))
In simple words: We changed \( \sin^{-1}x \) to \( t \). Then we found the derivative. After substituting, we integrated \( t \) to get \( \frac{t^2}{2} \), and then put \( \sin^{-1}x \) back for \( t \).

Exam Tip: For integrals involving inverse trigonometric functions, always check if the derivative of the inverse function is present in the integrand to use substitution effectively.

Question 24. \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\)
Answer: The initial integral is given as I = \(\int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} dx\). First, we simplify the denominator of the given expression to I = \(\int \frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)} dx\). We then assume that \( 3 \cos x+2 \sin x \) is equal to \( t \). Differentiating this, we find that \( (-3 \sin x + 2 \cos x) dx \) becomes \( dt \). Substituting these into the integral changes it to I = \(\int \frac{dt}{2t}\). We can take out the constant \( \frac{1}{2} \), leaving \( \frac{1}{2} \int \frac{1}{t} dt \). The integration of \( \frac{1}{t} \) yields \( \log |t| \), so the result is \( \frac{1}{2} \log |t| + c \). Finally, we replace \( t \) with \( (3 \cos x + 2 \sin x) \) to get the solution. The final answer is \( \frac{1}{2} \log |3 \cos x + 2 \sin x| + c \). (since \( t = 3 \cos x + 2 \sin x \))
In simple words: We made the denominator \( t \), and then the numerator became part of \( dt \). This simplified the integral into a basic \( \frac{1}{t} \) form. Then we just integrated it and put the original expression back.

Exam Tip: When the numerator is a multiple of the derivative of the denominator, substitute the denominator with 't'. This often simplifies the integral to \( \int \frac{1}{t} dt \).

Question 25. \(\frac{1}{\cos ^2 x(1-\tan x)^2}\)
Answer: We start with the given integral I = \(\int \frac{1}{\cos ^2 x(1-\tan x)^2} dx\). First, we transform \( \frac{1}{\cos^2 x} \) into \( \sec^2 x \) to simplify the expression, so I = \(\int \frac{\sec ^2 x}{(1-\tan x)^2} dx\). Next, we let \( 1-\tan x \) be equal to \( t \). Differentiating this choice, we find that \( -\sec^2 x dx \) becomes \( dt \). This means \( \sec^2 x dx \) is equivalent to \( -dt \). Substituting these into the integral gives us I = \(\int \frac{-dt}{t^2}\). We can rewrite this as \( -\int t^{-2} dt \). Upon integration, we achieve \( -\frac{t^{-2+1}}{-2+1} + c \). Simplifying this expression results in \( \frac{1}{t} + c \). Finally, substituting \( t \) back with \( 1-\tan x \) yields the complete solution. The answer is \( \frac{1}{1-\tan x} + c \). (since \( t = 1-\tan x \))
In simple words: We converted \( \frac{1}{\cos^2 x} \) to \( \sec^2 x \). Then we let \( (1-\tan x) \) be \( t \). The derivative of \( t \) appeared, making the integral easy to solve as \( -\int t^{-2} dt \).

Exam Tip: Recognizing \( \frac{1}{\cos^2 x} \) as \( \sec^2 x \) is key here. Also, remember that the derivative of \( \tan x \) is \( \sec^2 x \).

Question 26. \(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
Answer: We consider the integral I = \(\int \frac{\cos \sqrt{x}}{\sqrt{x}} dx\). Let's assume that \( \sqrt{x} \) is equal to \( t \). Differentiating both sides, we get \( \frac{1}{2\sqrt{x}} dx = dt \). This further implies that \( \frac{1}{\sqrt{x}} dx \) is equivalent to \( 2dt \). Substituting these into the integral transforms it into I = \(\int \cos t \cdot 2dt\). We can take out the constant 2, making it \( 2 \int \cos t dt \). Integrating \( \cos t \) gives \( \sin t \), so we have \( 2 \sin t + c \). Finally, replacing \( t \) with \( \sqrt{x} \) gives the answer. The solution is \( 2 \sin \sqrt{x} + c \). (since \( t = \sqrt{x} \))
In simple words: We let \( \sqrt{x} \) be \( t \). Its derivative, \( \frac{1}{\sqrt{x}} dx \), was already in the integral. This changed the problem to integrating \( \cos t \), which is straightforward.

Exam Tip: When \( \sqrt{x} \) is inside another function, consider substituting \( t = \sqrt{x} \). Remember that \( \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \).

Question 27. \(\sqrt{\sin 2 x} \cos 2x\)
Answer: We begin with the integral I = \(\int \sqrt{\sin 2 x} \cos 2x dx\). Let's assume \( \sin 2x \) is equal to \( t \). Differentiating this gives \( 2 \cos 2x dx = dt \). This means \( \cos 2x dx \) can be replaced by \( \frac{dt}{2} \). Substituting these into the integral changes it to I = \(\int \sqrt{t} \frac{dt}{2}\). We pull out the constant \( \frac{1}{2} \), getting \( \frac{1}{2} \int t^{\frac{1}{2}} dt \). Integrating \( t^{\frac{1}{2}} \) results in \( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \), so we have \( \frac{1}{2} \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c \). Simplifying the expression, this becomes \( \frac{1}{3} t^{\frac{3}{2}} + c \). Finally, substituting \( t \) back with \( \sin 2x \) gives the solution. The solution is \( \frac{1}{3} (\sin 2x)^{\frac{3}{2}} + c \). (since \( t = \sin 2x \))
In simple words: We set \( \sin 2x \) as \( t \). Its derivative \( \cos 2x dx \) was present, so we easily replaced it with \( dt \). This changed the problem to integrating \( \sqrt{t} \).

Exam Tip: Look for a function and its derivative in the integrand. If \( f(x) \) is present, and \( f'(x) dx \) is also present (or a multiple of it), substitution with \( u = f(x) \) is generally the best approach.

Question 28. \(\frac{\cos x}{\sqrt{1+\sin x}}\)
Answer: We start with the given integral I = \(\int \frac{\cos x}{\sqrt{1+\sin x}} dx\). Let's assume that \( 1+\sin x \) is equal to \( t \). Differentiating this choice, we obtain \( \cos x dx = dt \). Substituting these into the integral transforms it to I = \(\int \frac{dt}{\sqrt{t}}\). We can write this as \( \int t^{-\frac{1}{2}} dt \). Upon integration, we achieve \( \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + c \). Simplifying this expression gives us \( 2 t^{\frac{1}{2}} + c \). This is equivalent to \( 2 \sqrt{t} + c \). Finally, replacing \( t \) with \( 1+\sin x \) gives the complete solution. The solution is \( 2 \sqrt{1+\sin x} + c \). (since \( t = 1+\sin x \))
In simple words: We made \( 1+\sin x \) equal to \( t \). Its derivative \( \cos x dx \) was conveniently in the numerator. This let us integrate \( \frac{1}{\sqrt{t}} \) easily, then substitute back.

Exam Tip: When a function is inside a square root in the denominator and its derivative is in the numerator, substituting the expression inside the root is often the correct method.

Question 29. cot x log sin x
Answer: We begin with the integral I = \(\int \cot x \log \sin x dx\). Let's assume that \( \log \sin x \) is equal to \( t \). Differentiating this, we get \( \frac{1}{\sin x} \cos x dx = dt \). This simplifies to \( \cot x dx = dt \). Substituting these into the integral yields I = \(\int t dt\). Integrating this expression gives \( \frac{t^2}{2} + c \). Finally, replacing \( t \) with \( \log \sin x \) results in the complete solution. The answer is \( \frac{(\log \sin x)^2}{2} + c \). (since \( t = \log \sin x \))
In simple words: We set \( \log \sin x \) as \( t \). Its derivative turned out to be \( \cot x dx \), which was already in the problem. This made the integral a simple \( \int t dt \).

Exam Tip: Remember that the derivative of \( \log|f(x)| \) is \( \frac{f'(x)}{f(x)} \). This identity is frequently useful for substitution in integrals.

Question 30. \(\frac{\sin x}{1+\cos x}\)
Answer: We consider the integral I = \(\int \frac{\sin x}{1+\cos x}dx\). Let's assume that \( 1+\cos x \) is equal to \( t \). Differentiating this, we get \( -\sin x dx = dt \). This means \( \sin x dx \) can be replaced by \( -dt \). Substituting these into the integral transforms it to I = \(\int \frac{-dt}{t}\). We can take out the negative sign, giving \( -\int \frac{1}{t} dt \). Integrating \( \frac{1}{t} \) yields \( \log |t| \), so we have \( -\log |t| + c \). Finally, replacing \( t \) with \( 1+\cos x \) gives the complete solution. The answer is \( -\log |1+\cos x| + c \).
In simple words: We let \( 1+\cos x \) be \( t \). Its derivative \( -\sin x dx \) was available in the numerator. This transformed the integral into a straightforward \( -\int \frac{1}{t} dt \).

Exam Tip: Always look for functions whose derivative is also present in the integrand. The derivative of \( 1+\cos x \) is \( -\sin x \), which is a clear sign for substitution.

Question 31. \(\frac{\sin x}{(1+\cos x)^2}\)
Answer: We start with the integral I = \(\int \frac{\sin x}{(1+\cos x)^2} dx\). Let's assume that \( 1+\cos x \) is equal to \( t \). Differentiating this, we get \( -\sin x dx = dt \). This means \( \sin x dx \) can be replaced by \( -dt \). Substituting these into the integral transforms it to I = \(\int \frac{-dt}{t^2}\). We can write this as \( -\int t^{-2} dt \). Upon integration, we get \( -\frac{t^{-2+1}}{-2+1} + c \). Simplifying the expression results in \( \frac{1}{t} + c \). Finally, replacing \( t \) with \( 1+\cos x \) gives the complete solution. The answer is \( \frac{1}{1+\cos x} + c \). (since \( t = 1+\cos x \))
In simple words: We used \( t = 1+\cos x \). This made \( \sin x dx \) turn into \( -dt \). The integral then became \( -\int t^{-2} dt \), which we could easily solve.

Exam Tip: When the denominator is a power of a function and its derivative (or a multiple) is in the numerator, substituting the base of the power in the denominator with 't' is often helpful.

Question 32. \(\frac{1}{1+\cot x}\)
Answer: We begin with the integral I = \(\int \frac{1}{1+\cot x} dx\). First, we rewrite \( \cot x \) as \( \frac{\cos x}{\sin x} \) in the denominator. Then, we simplify the fraction to I = \(\int \frac{\sin x}{\sin x+\cos x} dx\). To make the numerator resemble the derivative of the denominator, we multiply and divide by 2. We then rearrange the numerator as \( (\sin x+\cos x) - (\cos x - \sin x) \). This allows us to split the integral into two parts: I = \( \frac{1}{2} \int \left(1 - \frac{\cos x - \sin x}{\sin x+\cos x}\right) dx \). Let's assume that \( \sin x+\cos x \) is equal to \( t \). Differentiating this, we find \( (\cos x - \sin x) dx = dt \). Substituting these into the integral gives I = \( \frac{1}{2} \left(x - \int \frac{dt}{t}\right) \). Integrating \( \frac{1}{t} \) gives \( \log |t| \), so we get \( \frac{1}{2} (x - \log |t|) + c \). Finally, replacing \( t \) with \( \sin x+\cos x \) gives the complete solution. The answer is \( \frac{1}{2} (x - \log |\sin x+\cos x|) + c \). (since \( t = \sin x+\cos x \))
In simple words: We converted \( \cot x \) to \( \frac{\cos x}{\sin x} \), then manipulated the fraction by adding and subtracting terms in the numerator. This split the integral into two simpler parts, one of which was easily solved by substitution.

Exam Tip: For integrals of the form \( \frac{1}{1+\tan x} \) or \( \frac{1}{1+\cot x} \), a common strategy is to convert \( \tan x \) or \( \cot x \) to \( \sin x \) and \( \cos x \), then manipulate the numerator to create terms related to the denominator and its derivative.

Question 33. \(\frac{1}{1-\tan x}\)
Answer: We begin with the integral I = \(\int \frac{1}{1-\tan x} dx\). First, we rewrite \( \tan x \) as \( \frac{\sin x}{\cos x} \) in the denominator. Then, we simplify the fraction to I = \(\int \frac{\cos x}{\cos x - \sin x} dx\). To prepare the numerator for differentiation, we multiply and divide by 2. We then rearrange the numerator as \( (\cos x - \sin x) + (\cos x + \sin x) \). This allows us to split the integral into two parts: I = \( \frac{1}{2} \int \left(1 + \frac{\cos x + \sin x}{\cos x - \sin x}\right) dx \). Let's assume that \( \cos x - \sin x \) is equal to \( t \). Differentiating this, we find \( (-\sin x - \cos x) dx = dt \). This implies \( (\sin x + \cos x) dx = -dt \). Substituting these into the integral gives I = \( \frac{1}{2} \left(x - \int \frac{dt}{t}\right) \). Integrating \( \frac{1}{t} \) gives \( \log |t| \), so we get \( \frac{1}{2} (x - \log |t|) + c \). Finally, replacing \( t \) with \( \cos x - \sin x \) gives the complete solution. The answer is \( \frac{1}{2} (x - \log |\cos x - \sin x|) + c \). (since \( t = \cos x - \sin x \))
In simple words: We changed \( \tan x \) to \( \frac{\sin x}{\cos x} \), then broke the integral into two parts. One part was easy, and the other used substitution by setting \( (\cos x - \sin x) \) as \( t \).

Exam Tip: Similar to \( \frac{1}{1+\cot x} \), for \( \frac{1}{1-\tan x} \), convert \( \tan x \) to \( \sin x \) and \( \cos x \), then adjust the numerator to be the sum of the denominator and its derivative, or a similar algebraic manipulation.

Question 34. \(\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}\)
Answer: We begin with the integral I = \(\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} dx\). First, we adjust the denominator by dividing and multiplying by \( \cos x \). This transforms the expression into I = \(\int \frac{\sqrt{\tan x}}{\tan x \cdot \cos^2 x} dx\). We can simplify this further to I = \(\int \frac{1}{\sqrt{\tan x}} \sec^2 x dx\). Let's assume that \( \tan x \) is equal to \( t \). Differentiating this, we get \( \sec^2 x dx = dt \). Substituting these into the integral transforms it to I = \(\int \frac{1}{\sqrt{t}} dt\). We can write this as \( \int t^{-\frac{1}{2}} dt \). Upon integration, we obtain \( \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + c \). Simplifying this expression gives \( 2 t^{\frac{1}{2}} + c \). This is equivalent to \( 2 \sqrt{t} + c \). Finally, replacing \( t \) with \( \tan x \) gives the complete solution. The answer is \( 2 \sqrt{\tan x} + c \). (since \( t = \tan x \))
In simple words: We changed the denominator to include \( \tan x \) and \( \sec^2 x \). Then we let \( \tan x \) be \( t \), which made the integral a simple power rule problem.

Exam Tip: When \( \tan x \) and \( \sin x \cos x \) are present, try to manipulate the expression to get \( \tan x \) and \( \sec^2 x \), as \( \sec^2 x \) is the derivative of \( \tan x \).

Question 35. \(\frac{(1+\log x)^2}{x}\)
Answer: We consider the integral I = \(\int \frac{(1+\log x)^2}{x} dx\). Let's assume that \( 1+\log x \) is equal to \( t \). Differentiating this, we get \( \frac{1}{x} dx = dt \). Substituting these into the integral transforms it into I = \(\int t^2 dt\). Integrating this expression gives \( \frac{t^3}{3} + c \). Finally, replacing \( t \) with \( 1+\log x \) results in the complete solution. The answer is \( \frac{(1+\log x)^3}{3} + c \). (since \( t = 1+\log x \))
In simple words: We set \( 1+\log x \) as \( t \). Its derivative, \( \frac{1}{x} dx \), was already in the problem, making the integral a basic \( \int t^2 dt \).

Exam Tip: Always recognize that \( \frac{1}{x} \) is the derivative of \( \log x \). This is a strong indicator for substituting \( 1+\log x \) as 't' when it appears in a complex expression.

Question 36. \(\frac{(x+1)(x+\log x)^2}{x}\)
Answer: We begin with the integral I = \(\int \frac{(x+1)(x+\log x)^2}{x} dx\). First, we separate the fraction in the first term of the numerator, I = \(\int \left(\frac{x}{x}+\frac{1}{x}\right) (x+\log x)^2 dx\). This simplifies the expression to I = \(\int \left(1+\frac{1}{x}\right) (x+\log x)^2 dx\). Let's assume that \( x+\log x \) is equal to \( t \). Differentiating this, we get \( \left(1+\frac{1}{x}\right) dx = dt \). Substituting these into the integral transforms it into I = \(\int t^2 dt\). Integrating this expression gives \( \frac{t^3}{3} + c \). Finally, replacing \( t \) with \( x+\log x \) results in the complete solution. The answer is \( \frac{(x+\log x)^3}{3} + c \). (since \( t = x+\log x \))
In simple words: We split \( \frac{x+1}{x} \) into \( 1+\frac{1}{x} \). Then, by letting \( x+\log x \) be \( t \), its derivative \( (1+\frac{1}{x}) dx \) was perfectly matched, simplifying the integral to \( \int t^2 dt \).

Exam Tip: Look for opportunities to algebraically simplify the integrand before applying substitution. Recognizing \( 1+\frac{1}{x} \) as the derivative of \( x+\log x \) is crucial here.

Question 37. \(\frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8}\)
Answer: We begin with the integral I = \(\int \frac{x^3 \sin (\tan^{-1} x^4)}{1+x^8} dx\). Let's assume that \( \tan^{-1} x^4 \) is equal to \( t \). Differentiating this, we get \( \frac{1}{1+(x^4)^2} \cdot 4x^3 dx = dt \). This simplifies to \( \frac{4x^3}{1+x^8} dx = dt \). Further rearrangement yields \( \frac{x^3}{1+x^8} dx = \frac{dt}{4} \). Substituting these into the integral transforms it into I = \(\int \sin t \frac{dt}{4}\). We can take out the constant \( \frac{1}{4} \), getting \( \frac{1}{4} \int \sin t dt \). Integrating \( \sin t \) results in \( -\cos t \), so we have \( -\frac{1}{4} \cos t + c \). Finally, replacing \( t \) with \( \tan^{-1} x^4 \) gives the complete solution. The answer is \( -\frac{1}{4} \cos (\tan^{-1} x^4) + c \). (since \( t = \tan^{-1} x^4 \))
In simple words: We let \( \tan^{-1} x^4 \) be \( t \). Its derivative, which includes \( \frac{x^3}{1+x^8} \), allowed us to convert the integral to a simple \( \int \sin t dt \) form.

Exam Tip: For complex inverse trigonometric functions within an integrand, consider substituting the inverse function itself. Remember the chain rule for derivatives, especially for \( \tan^{-1}(u) \).

Question 38. \(\int \frac{\left(10 x^9+10^x \log _{\mathrm{e}} 10\right) d x}{x^{10}+10^x} = \ldots\)
(A) \(10^x – x^{10} + C\)
(B) \(10^x + x^{10} + C\)
(C) \((10^x – x^{10})^{-1} + c\)
(D) \(\log(10^x + x^{10}) + c\)
Answer: (D) \(\log(10^x + x^{10}) + c\)
In simple words: We set the denominator \( (x^{10}+10^x) \) as \( t \). The numerator then became exactly \( dt \), which simplifies the integral to \( \int \frac{1}{t} dt \), resulting in \( \log|t| \).

Exam Tip: This integral is of the form \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C \). Always look for cases where the numerator is the derivative of the denominator.

Question 39. \(\int \frac{d x}{\sin ^2 x \cos ^2 x} = \ldots\)
(B) \(\tan x - \cot x + c\)
(c) \(\tan x \cot x + c\)
(D) \(\tan x - \cot 2x + c\)
Answer: (B) \(\tan x - \cot x + c\)
In simple words: We used the identity \( \sin^2 x + \cos^2 x = 1 \) in the numerator. This let us split the fraction into two terms, \( \frac{1}{\cos^2 x} \) and \( \frac{1}{\sin^2 x} \), which simplify to \( \sec^2 x \) and \( \operatorname{cosec}^2 x \), both of which are easy to integrate.

Exam Tip: When \( 1 \) is in the numerator and the denominator has products of \( \sin x \) and \( \cos x \), using the identity \( 1 = \sin^2 x + \cos^2 x \) is a common and effective strategy.