GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.10

Evaluate the following integrals:

Question 1. \( \int_{0}^{1} \frac{x}{x^{2}+1} dx \)
Answer: Let \( I = \int_{0}^{1} \frac{x}{x^{2}+1} dx \).
Put \( x^2 + 1 = t \). Then \( 2x \, dx = dt \).
When \( x = 1 \), \( t = 1^2 + 1 = 2 \).
When \( x = 0 \), \( t = 0^2 + 1 = 1 \).
\( \therefore I = \int_{1}^{2} \frac{1}{t} \frac{dt}{2} \)
\( = \frac{1}{2} \int_{1}^{2} \frac{dt}{t} \)
\( = \frac{1}{2} [\log |t|]_{1}^{2} \)
\( = \frac{1}{2} (\log 2 - \log 1) \)
\( = \frac{1}{2} \log 2 \).
In simple words: We substitute \( x^2+1 \) with \( t \) to simplify the integral. After changing the limits and integrating, we get half of the natural logarithm of 2.

Exam Tip: Remember to change the limits of integration when you perform a substitution (u-substitution) in a definite integral.

Question 2. \( \int_{0}^{\frac{\pi}{2}} \sqrt{\sin\phi}\cos^5\phi d\phi \)
Answer: Let \( I = \int_{0}^{\frac{\pi}{2}} \sqrt{\sin\phi}\cos^5\phi d\phi \).
Rewrite \( \cos^5\phi \) as \( \cos^4\phi \cdot \cos\phi \).
\( = \int_{0}^{\frac{\pi}{2}} \sqrt{\sin\phi} (1-\sin^2\phi)^2 \cos\phi d\phi \).
Put \( \sin\phi = t \). Then \( \cos\phi \, d\phi = dt \).
When \( \phi = 0 \), \( t = \sin 0 = 0 \).
When \( \phi = \frac{\pi}{2} \), \( t = \sin \frac{\pi}{2} = 1 \).
\( \therefore I = \int_{0}^{1} \sqrt{t} (1-t^2)^2 dt \)
\( = \int_{0}^{1} t^{\frac{1}{2}} (1 - 2t^2 + t^4) dt \)
\( = \int_{0}^{1} (t^{\frac{1}{2}} - 2t^{\frac{5}{2}} + t^{\frac{9}{2}}) dt \)
\( = \left[ \frac{t^{\frac{3}{2}}}{\frac{3}{2}} - 2 \frac{t^{\frac{7}{2}}}{\frac{7}{2}} + \frac{t^{\frac{11}{2}}}{\frac{11}{2}} \right]_{0}^{1} \)
\( = \left[ \frac{2}{3} t^{\frac{3}{2}} - \frac{4}{7} t^{\frac{7}{2}} + \frac{2}{11} t^{\frac{11}{2}} \right]_{0}^{1} \)
\( = \left( \frac{2}{3} - \frac{4}{7} + \frac{2}{11} \right) - (0) \)
\( = \frac{2 \cdot 7 \cdot 11 - 4 \cdot 3 \cdot 11 + 2 \cdot 3 \cdot 7}{3 \cdot 7 \cdot 11} \)
\( = \frac{154 - 132 + 42}{231} \)
\( = \frac{64}{231} \).
In simple words: We rewrite the cosine term, then use a substitution \( \sin\phi = t \). This changes the integral into a polynomial in \( t \), which is easier to integrate. After evaluating at the limits, the result is \( \frac{64}{231} \).

Exam Tip: For integrals involving powers of sine and cosine, look for opportunities to use the identity \( \sin^2x + \cos^2x = 1 \) to reduce one power to a single term, allowing for a u-substitution.

Question 3. \( \int_{0}^{1} \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx \)
Answer: Let \( I = \int_{0}^{1} \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx \).
We know the identity: \( \sin^{-1}\left(\frac{2x}{1+x^{2}}\right) = 2 \tan^{-1}x \).
\( \therefore I = \int_{0}^{1} 2 \tan^{-1}x \, dx \).
We will use integration by parts, taking \( \tan^{-1}x \) as the first function and \( 1 \) as the second function.
Formula for integration by parts: \( \int u \, dv = uv - \int v \, du \).
Let \( u = \tan^{-1}x \implies du = \frac{1}{1+x^2} dx \).
Let \( dv = dx \implies v = x \).
\( I = 2 \left[ [x \tan^{-1}x]_{0}^{1} - \int_{0}^{1} \frac{x}{1+x^2} dx \right] \)
For the integral \( \int_{0}^{1} \frac{x}{1+x^2} dx \), let \( w = 1+x^2 \), so \( dw = 2x \, dx \).
When \( x=0, w=1 \); when \( x=1, w=2 \).
\( \int_{0}^{1} \frac{x}{1+x^2} dx = \int_{1}^{2} \frac{1}{w} \frac{dw}{2} = \frac{1}{2} [\log|w|]_{1}^{2} = \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} \log 2 \).
Now substitute back into \( I \):
\( I = 2 \left[ (1 \cdot \tan^{-1}1 - 0 \cdot \tan^{-1}0) - \frac{1}{2} \log 2 \right] \)
\( I = 2 \left[ (\frac{\pi}{4} - 0) - \frac{1}{2} \log 2 \right] \)
\( I = 2 \left( \frac{\pi}{4} - \frac{1}{2} \log 2 \right) \)
\( I = \frac{\pi}{2} - \log 2 \).
In simple words: First, we use a known trigonometric identity to simplify the inverse sine term. Then, we apply integration by parts to solve the new integral. Finally, we evaluate the expression at the given limits to find the numerical answer.

Exam Tip: Recognizing standard trigonometric substitutions and identities, like \( \sin^{-1}\left(\frac{2x}{1+x^{2}}\right) = 2 \tan^{-1}x \), can greatly simplify integration problems. Also, remember the steps for integration by parts carefully.

Question 4. \( \int_{0}^{2}x\sqrt{x+2} dx \)
Answer: Let \( I = \int_{0}^{2}x\sqrt{x+2} dx \).
Put \( t^2 = x+2 \). Then \( x = t^2-2 \).
Differentiating both sides: \( 2t \, dt = dx \).
When \( x = 0 \), \( t^2 = 0+2 \implies t = \sqrt{2} \).
When \( x = 2 \), \( t^2 = 2+2 \implies t = \sqrt{4} \implies t = 2 \).
Substitute these into the integral:
\( I = \int_{\sqrt{2}}^{2} (t^2-2) \sqrt{t^2} (2t \, dt) \)
\( = \int_{\sqrt{2}}^{2} (t^2-2) \cdot t \cdot 2t \, dt \)
\( = \int_{\sqrt{2}}^{2} 2t^2 (t^2-2) dt \)
\( = \int_{\sqrt{2}}^{2} (2t^4 - 4t^2) dt \)
\( = \left[ \frac{2t^5}{5} - \frac{4t^3}{3} \right]_{\sqrt{2}}^{2} \)
\( = \left( \frac{2(2)^5}{5} - \frac{4(2)^3}{3} \right) - \left( \frac{2(\sqrt{2})^5}{5} - \frac{4(\sqrt{2})^3}{3} \right) \)
\( = \left( \frac{2 \cdot 32}{5} - \frac{4 \cdot 8}{3} \right) - \left( \frac{2 \cdot 4\sqrt{2}}{5} - \frac{4 \cdot 2\sqrt{2}}{3} \right) \)
\( = \left( \frac{64}{5} - \frac{32}{3} \right) - \left( \frac{8\sqrt{2}}{5} - \frac{8\sqrt{2}}{3} \right) \)
\( = \frac{64 \cdot 3 - 32 \cdot 5}{15} - \frac{8\sqrt{2} \cdot 3 - 8\sqrt{2} \cdot 5}{15} \)
\( = \frac{192 - 160}{15} - \frac{24\sqrt{2} - 40\sqrt{2}}{15} \)
\( = \frac{32}{15} - \frac{-16\sqrt{2}}{15} \)
\( = \frac{32 + 16\sqrt{2}}{15} \)
\( = \frac{16(2+\sqrt{2})}{15} \).
In simple words: We used a substitution \( t^2 = x+2 \) to simplify the square root term. This transformed the integral into a polynomial, which is straightforward to integrate. We then evaluated the resulting expression at the new limits.

Exam Tip: When a square root is present, a substitution of the form \( u^2 = \text{expression under root} \) often helps eliminate the square root and simplifies the integral.

Question 5. \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} dx \)
Answer: Let \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} dx \).
Put \( \cos x = t \).
Differentiating both sides: \( -\sin x \, dx = dt \implies \sin x \, dx = -dt \).
When \( x = 0 \), \( t = \cos 0 = 1 \).
When \( x = \frac{\pi}{2} \), \( t = \cos \frac{\pi}{2} = 0 \).
Substitute these values into the integral:
\( I = \int_{1}^{0} \frac{-dt}{1+t^2} \)
\( = - \int_{1}^{0} \frac{1}{1+t^2} dt \)
\( = \int_{0}^{1} \frac{1}{1+t^2} dt \) (changing the limits swaps the sign)
\( = [\tan^{-1}t]_{0}^{1} \)
\( = \tan^{-1}1 - \tan^{-1}0 \)
\( = \frac{\pi}{4} - 0 \)
\( = \frac{\pi}{4} \).
In simple words: We substitute \( \cos x \) with \( t \) to simplify the expression. This converts the integral into a standard form whose antiderivative is \( \tan^{-1}t \). Evaluating this at the new limits gives the final answer.

Exam Tip: Look for a term whose derivative is also present (or a multiple of it) for a simple u-substitution. Also, remember that switching the limits of integration changes the sign of the definite integral.

Question 6. \( \int_{0}^{2} \frac{d x}{x+4-x^{2}} \)
Answer: Let \( I = \int_{0}^{2} \frac{d x}{x+4-x^{2}} \).
First, complete the square in the denominator:
\( x+4-x^2 = -(x^2-x-4) \)
\( = -\left( x^2 - x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - 4 \right) \)
\( = -\left( \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} - 4 \right) \)
\( = -\left( \left(x - \frac{1}{2}\right)^2 - \frac{1+16}{4} \right) \)
\( = -\left( \left(x - \frac{1}{2}\right)^2 - \frac{17}{4} \right) \)
\( = \frac{17}{4} - \left(x - \frac{1}{2}\right)^2 \).
So, \( I = \int_{0}^{2} \frac{d x}{\frac{17}{4} - \left(x - \frac{1}{2}\right)^2} \).
This is of the form \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \).
Here, \( a^2 = \frac{17}{4} \implies a = \frac{\sqrt{17}}{2} \).
Let \( u = x - \frac{1}{2} \). Then \( du = dx \).
When \( x=0, u = 0 - \frac{1}{2} = -\frac{1}{2} \).
When \( x=2, u = 2 - \frac{1}{2} = \frac{3}{2} \).
\( I = \int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d u}{\left(\frac{\sqrt{17}}{2}\right)^2 - u^2} \)
\( = \frac{1}{2 \cdot \frac{\sqrt{17}}{2}} \left[ \log \left| \frac{\frac{\sqrt{17}}{2} + u}{\frac{\sqrt{17}}{2} - u} \right| \right]_{-\frac{1}{2}}^{\frac{3}{2}} \)
\( = \frac{1}{\sqrt{17}} \left[ \log \left| \frac{\sqrt{17} + 2u}{\sqrt{17} - 2u} \right| \right]_{-\frac{1}{2}}^{\frac{3}{2}} \)
\( = \frac{1}{\sqrt{17}} \left( \log \left| \frac{\sqrt{17} + 2(\frac{3}{2})}{\sqrt{17} - 2(\frac{3}{2})} \right| - \log \left| \frac{\sqrt{17} + 2(-\frac{1}{2})}{\sqrt{17} - 2(-\frac{1}{2})} \right| \right) \)
\( = \frac{1}{\sqrt{17}} \left( \log \left| \frac{\sqrt{17} + 3}{\sqrt{17} - 3} \right| - \log \left| \frac{\sqrt{17} - 1}{\sqrt{17} + 1} \right| \right) \)
Using \( \log A - \log B = \log \frac{A}{B} \):
\( = \frac{1}{\sqrt{17}} \log \left| \frac{\frac{\sqrt{17} + 3}{\sqrt{17} - 3}}{\frac{\sqrt{17} - 1}{\sqrt{17} + 1}} \right| \)
\( = \frac{1}{\sqrt{17}} \log \left| \frac{(\sqrt{17} + 3)(\sqrt{17} + 1)}{(\sqrt{17} - 3)(\sqrt{17} - 1)} \right| \)
\( = \frac{1}{\sqrt{17}} \log \left| \frac{17 + \sqrt{17} + 3\sqrt{17} + 3}{17 - \sqrt{17} - 3\sqrt{17} + 3} \right| \)
\( = \frac{1}{\sqrt{17}} \log \left| \frac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}} \right| \)
\( = \frac{1}{\sqrt{17}} \log \left| \frac{4(5 + \sqrt{17})}{4(5 - \sqrt{17})} \right| \)
\( = \frac{1}{\sqrt{17}} \log \left| \frac{5 + \sqrt{17}}{5 - \sqrt{17}} \right| \).
To simplify further, rationalize the argument of the logarithm:
\( = \frac{1}{\sqrt{17}} \log \left| \frac{5 + \sqrt{17}}{5 - \sqrt{17}} \cdot \frac{5 + \sqrt{17}}{5 + \sqrt{17}} \right| \)
\( = \frac{1}{\sqrt{17}} \log \left| \frac{(5 + \sqrt{17})^2}{25 - 17} \right| \)
\( = \frac{1}{\sqrt{17}} \log \left| \frac{25 + 17 + 10\sqrt{17}}{8} \right| \)
\( = \frac{1}{\sqrt{17}} \log \left| \frac{42 + 10\sqrt{17}}{8} \right| \)
\( = \frac{1}{\sqrt{17}} \log \left| \frac{21 + 5\sqrt{17}}{4} \right| \).
In simple words: First, we complete the square in the denominator to transform the integral into a standard form. Then, we apply the known integration formula for \( \frac{1}{a^2-x^2} \) and evaluate it at the given limits. Finally, we simplify the logarithmic expression by rationalizing its argument.

Exam Tip: When integrating rational functions involving quadratic denominators, completing the square is often the first step. Recognize standard integral forms like \( \int \frac{dx}{a^2-x^2} \) or \( \int \frac{dx}{x^2+a^2} \).

Question 7. \( \int_{-1}^{1} \frac{d x}{x^{2}+2x+5} \)
Answer: Let \( I = \int_{-1}^{1} \frac{d x}{x^{2}+2x+5} \).
First, complete the square in the denominator:
\( x^2+2x+5 = (x^2+2x+1) + 4 = (x+1)^2 + 4 \).
So, \( I = \int_{-1}^{1} \frac{d x}{(x+1)^2 + 2^2} \).
This is of the form \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \).
Here, \( a = 2 \). Let \( u = x+1 \). Then \( du = dx \).
When \( x=-1, u = -1+1 = 0 \).
When \( x=1, u = 1+1 = 2 \).
\( I = \int_{0}^{2} \frac{d u}{u^2 + 2^2} \)
\( = \frac{1}{2} \left[ \tan^{-1}\left(\frac{u}{2}\right) \right]_{0}^{2} \)
\( = \frac{1}{2} \left( \tan^{-1}\left(\frac{2}{2}\right) - \tan^{-1}\left(\frac{0}{2}\right) \right) \)
\( = \frac{1}{2} (\tan^{-1}1 - \tan^{-1}0) \)
\( = \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) \)
\( = \frac{\pi}{8} \).
In simple words: We first rewrite the denominator by completing the square, turning it into a form that fits a standard integration formula. After that, we perform the integration and substitute the limits to get the final numerical result.

Exam Tip: For definite integrals with quadratic denominators, completing the square is a key technique. Remember the integral forms for \( \frac{1}{x^2+a^2} \) and \( \frac{1}{a^2-x^2} \).

Question 8. \( \int_{1}^{2} \left(\frac{1}{x} – \frac{1}{2 x^{2}}\right) e^{2x} dx \)
Answer: Let \( I = \int_{1}^{2} e^{2x} \left(\frac{1}{x} – \frac{1}{2 x^{2}}\right) dx \).
We know the formula: \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \).
A variant is \( \int e^{ax} (f(x) + \frac{1}{a}f'(x)) dx = e^{ax} \frac{f(x)}{a} + C \).
Here, compare \( \left(\frac{1}{x} – \frac{1}{2 x^{2}}\right) \) with \( f(x) + \frac{1}{a}f'(x) \).
Let \( a=2 \). We need \( f(x) \) such that \( f(x) + \frac{1}{2}f'(x) = \frac{1}{x} – \frac{1}{2x^2} \).
If we choose \( f(x) = \frac{1}{x} \), then \( f'(x) = -\frac{1}{x^2} \).
Then \( f(x) + \frac{1}{2}f'(x) = \frac{1}{x} + \frac{1}{2}\left(-\frac{1}{x^2}\right) = \frac{1}{x} - \frac{1}{2x^2} \).
This matches our integrand. So, \( f(x) = \frac{1}{x} \).
Therefore, \( I = \left[ e^{2x} \frac{1}{2} \cdot \frac{1}{x} \right]_{1}^{2} \)
\( = \frac{1}{2} \left[ \frac{e^{2x}}{x} \right]_{1}^{2} \)
\( = \frac{1}{2} \left( \frac{e^{2 \cdot 2}}{2} - \frac{e^{2 \cdot 1}}{1} \right) \)
\( = \frac{1}{2} \left( \frac{e^4}{2} - e^2 \right) \)
\( = \frac{e^4}{4} - \frac{e^2}{2} \)
\( = \frac{e^2(e^2 - 2)}{4} \).
In simple words: This integral fits a special form where \( e^{ax} \) is multiplied by a function plus a scaled version of its derivative. By identifying \( f(x) = \frac{1}{x} \), we can use the direct formula for such integrals. Evaluating it at the given limits yields the result.

Exam Tip: Always be on the lookout for the integral form \( \int e^{ax}(f(x) + \frac{1}{a}f'(x))dx \). It often simplifies complex problems significantly. This identity is derived from integration by parts.

Choose the correct answers in questions 9 and 10:

Question 9. The value of the integral \( \int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} dx \) is
(A) 6
(B) 0
(C) 3
(D) 4
Answer: (A) 6
Solution: Let \( I = \int_{\frac{1}{3}}^{1} \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx \).
Factor out \( x^3 \) from the term \( (x-x^3) \):
\( (x-x^3)^{\frac{1}{3}} = [x^3(\frac{1}{x^2}-1)]^{\frac{1}{3}} = x (\frac{1}{x^2}-1)^{\frac{1}{3}} \).
Substitute this back into the integral:
\( I = \int_{\frac{1}{3}}^{1} \frac{x (\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^4} dx \)
\( = \int_{\frac{1}{3}}^{1} \frac{(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^3} dx \).
Put \( t = \frac{1}{x^2} - 1 \).
Then \( dt = -\frac{2}{x^3} dx \implies \frac{1}{x^3} dx = -\frac{1}{2} dt \).
When \( x = \frac{1}{3} \), \( t = \frac{1}{(\frac{1}{3})^2} - 1 = \frac{1}{\frac{1}{9}} - 1 = 9 - 1 = 8 \).
When \( x = 1 \), \( t = \frac{1}{1^2} - 1 = 1 - 1 = 0 \).
Substitute these into the integral:
\( I = \int_{8}^{0} t^{\frac{1}{3}} \left(-\frac{1}{2}\right) dt \)
\( = -\frac{1}{2} \int_{8}^{0} t^{\frac{1}{3}} dt \)
\( = \frac{1}{2} \int_{0}^{8} t^{\frac{1}{3}} dt \) (changing the limits swaps the sign)
\( = \frac{1}{2} \left[ \frac{t^{\frac{1}{3}+1}}{\frac{1}{3}+1} \right]_{0}^{8} \)
\( = \frac{1}{2} \left[ \frac{t^{\frac{4}{3}}}{\frac{4}{3}} \right]_{0}^{8} \)
\( = \frac{1}{2} \cdot \frac{3}{4} \left[ t^{\frac{4}{3}} \right]_{0}^{8} \)
\( = \frac{3}{8} (8^{\frac{4}{3}} - 0^{\frac{4}{3}}) \)
\( = \frac{3}{8} ((2^3)^{\frac{4}{3}}) \)
\( = \frac{3}{8} (2^4) \)
\( = \frac{3}{8} \cdot 16 \)
\( = 3 \cdot 2 \)
\( = 6 \).
In simple words: First, we simplify the term inside the cube root by factoring \( x^3 \). Then, we use a substitution \( t = \frac{1}{x^2} - 1 \) to transform the integral into a simpler power function of \( t \). After integrating and evaluating at the changed limits, the answer is 6.

Exam Tip: For complicated integrands, algebraic manipulation (like factoring) before substitution can make the problem much more manageable. Don't forget to adjust the integration limits when performing a substitution.

Question 10. If \( f(x) = \int_{0}^{x} t \sin t dt \), then \( f'(x) \) is
(A) \( \cos x + x \sin x \)
(B) \( x \sin x \)
(C) \( x \cos x \)
(D) \( \sin x + x \cos x \)
Answer: (B) \( x \sin x \)
Solution: We are given \( f(x) = \int_{0}^{x} t \sin t dt \).
To find \( f'(x) \), we use the Fundamental Theorem of Calculus, Part 1.
The theorem states that if \( F(x) = \int_{a}^{x} g(t) dt \), then \( F'(x) = g(x) \).
In this case, \( g(t) = t \sin t \).
So, \( f'(x) = x \sin x \).
Alternatively, we can first evaluate the integral using integration by parts:
Let \( u = t \implies du = dt \).
Let \( dv = \sin t \, dt \implies v = -\cos t \).
\( \int t \sin t \, dt = -t \cos t - \int (-\cos t) dt \)
\( = -t \cos t + \int \cos t \, dt \)
\( = -t \cos t + \sin t + C \).
Now apply the limits:
\( f(x) = [-t \cos t + \sin t]_{0}^{x} \)
\( = (-x \cos x + \sin x) - (-0 \cos 0 + \sin 0) \)
\( = -x \cos x + \sin x - 0 \)
\( = -x \cos x + \sin x \).
Then differentiate \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx} (-x \cos x + \sin x) \)
Use product rule for \( -x \cos x \): \( \frac{d}{dx}(uv) = u'v + uv' \).
\( \frac{d}{dx}(-x \cos x) = (-1)\cos x + (-x)(-\sin x) = -\cos x + x \sin x \).
And \( \frac{d}{dx}(\sin x) = \cos x \).
So, \( f'(x) = (-\cos x + x \sin x) + \cos x \)
\( = x \sin x \).
In simple words: The first method uses a fundamental rule of calculus that lets us directly find the derivative of an integral with a variable upper limit. The second method first solves the integral using integration by parts and then differentiates the result. Both methods give the same answer, \( x \sin x \).

Exam Tip: When faced with an integral that has a variable upper limit, applying the Fundamental Theorem of Calculus is usually the quickest way to find its derivative. It saves time compared to evaluating the integral first and then differentiating.