GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.1

Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Integrals Ex 7.1

Find an antiderivative (or integral) of the following by the method of inspection:

Question 1. sin 2x
Answer: We are aware that \( \frac{d}{dx} \cos 2x = -2\sin 2x \). We can also write this as \( \frac{d}{dx} \left( -\frac{1}{2} \cos 2x \right) = \sin 2x \). Therefore, an antiderivative of \( \sin 2x \) is \( -\frac{1}{2} \cos 2x + C \).
In simple words: We need to find a function that, when differentiated, results in \( \sin 2x \). That function is \( -\frac{1}{2} \cos 2x \), and we add C for the constant of integration.

Exam Tip: When using the inspection method, think about which function's derivative would yield the given function. Remember to adjust constant multipliers and add 'C'.

Question 2. cos 3x
Answer: We know that \( \frac{d}{dx} (\sin 3x) = 3 \cos 3x \). From this, we can deduce that \( \cos 3x = \frac{1}{3} \frac{d}{dx} (\sin 3x) \). This can also be expressed as \( \cos 3x = \frac{d}{dx} \left( \frac{1}{3} \sin 3x \right) \). Hence, an antiderivative of \( \cos 3x \) is \( \frac{1}{3} \sin 3x + C \).
In simple words: We are looking for a function whose derivative is \( \cos 3x \). The function \( \frac{1}{3} \sin 3x \) serves this purpose, and we include C as the constant of integration.

Exam Tip: For trigonometric functions like sine and cosine, recall their derivatives and use a constant factor to match the given expression.

Question 3. e^2x
Answer: We are aware that \( \frac{d}{dx}(e^{2x}) = 2e^{2x} \). This relationship allows us to write \( e^{2x} = \frac{1}{2} \frac{d}{dx} (e^{2x}) \). Consequently, we can say that \( e^{2x} = \frac{d}{dx} \left( \frac{1}{2} e^{2x} \right) \). Therefore, an antiderivative of \( e^{2x} \) is \( \frac{1}{2} e^{2x} + C \).
In simple words: To find the antiderivative of \( e^{2x} \), we consider that the derivative of \( e^{2x} \) is \( 2e^{2x} \). By dividing by 2, we get \( \frac{1}{2} e^{2x} \), which is the antiderivative, plus C.

Exam Tip: The derivative of \( e^{ax} \) is \( ae^{ax} \). To find the antiderivative of \( e^{ax} \), you will divide by 'a'.

Question 4. (ax + b)²
Answer: We understand that \( \frac{d}{dx} (ax + b)^3 = 3a(ax + b)^2 \). This expression helps us derive \( (ax + b)^2 = \frac{1}{3a} \frac{d}{dx}(ax + b)^3 \). We can also present this as \( (ax + b)^2 = \frac{d}{dx} \left[ \frac{1}{3a}(ax + b)^3 \right] \). Thus, an antiderivative of \( (ax + b)^2 \) is \( \frac{1}{3a}(ax + b)^3 + C \).
In simple words: We know that the power rule for derivatives involves \( (ax+b)^n \). The derivative of \( (ax+b)^3 \) is \( 3a(ax+b)^2 \). So, to get \( (ax+b)^2 \), we divide by \( 3a \) and add C.

Exam Tip: For functions of the form \( (ax+b)^n \), remember that the antiderivative is \( \frac{(ax+b)^{n+1}}{a(n+1)} + C \). Pay attention to the 'a' coefficient.

Question 5. sin 2x - 4e^3x
Answer: We recognise that \( \frac{d}{dx} (\cos 2x) = -2 \sin 2x \). This implies that \( \sin 2x = \frac{d}{dx} \left( -\frac{1}{2} \cos 2x \right) \). Also, for the exponential part, \( \frac{d}{dx} (e^{3x}) = 3e^{3x} \), which suggests \( 4e^{3x} = 4 \times \frac{1}{3} \frac{d}{dx} (e^{3x}) = \frac{d}{dx} \left( \frac{4}{3} e^{3x} \right) \). Therefore, an antiderivative of \( \sin 2x - 4e^{3x} \) is \( -\frac{1}{2} \cos 2x - \frac{4}{3} e^{3x} + C \).
In simple words: We find the antiderivative for each term separately. The antiderivative of \( \sin 2x \) is \( -\frac{1}{2} \cos 2x \). The antiderivative of \( 4e^{3x} \) is \( \frac{4}{3} e^{3x} \). Combine them and add the constant C.

Exam Tip: Antiderivatives can be found term by term. Apply the appropriate rules for each part of the expression.

Find the following integrals:

Question 6. \( \int(4e^{3x} + 1)dx \)
Answer: We can split the integral: \( \int(4e^{3x} + 1)dx = \int 4e^{3x} dx + \int 1 dx \). This simplifies to \( 4 \int e^{3x} dx + \int dx \). Integrating gives us \( 4 \left( \frac{e^{3x}}{3} \right) + x + C \). So the final answer is \( \frac{4}{3} e^{3x} + x + C \).
In simple words: Break the integral into two parts, one for \( 4e^{3x} \) and one for \( 1 \). Integrate each part separately. The integral of \( e^{3x} \) is \( \frac{e^{3x}}{3} \), and the integral of \( 1 \) is \( x \). Don't forget to add C.

Exam Tip: Always remember that the integral of a sum is the sum of the integrals. For \( e^{ax} \), the integral is \( \frac{e^{ax}}{a} \).

Question 7. \( \int x^2 \left( 1 - \frac{1}{x^2} \right)dx \)
Answer: First, expand the expression: \( \int x^2 \left( 1 - \frac{1}{x^2} \right)dx = \int (x^2 - 1)dx \). Now, integrate each term: \( \int x^2 dx - \int 1 dx \). Applying the power rule for integration, this becomes \( \frac{x^{2+1}}{2+1} - x + C \). Simplifying, we get \( \frac{x^3}{3} - x + C \).
In simple words: Multiply \( x^2 \) inside the bracket first. This makes the integral \( \int (x^2 - 1)dx \). Then, integrate \( x^2 \) to get \( \frac{x^3}{3} \) and integrate \( -1 \) to get \( -x \). Add C at the end.

Exam Tip: Simplify the integrand before integrating. The power rule states \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \) for \( n \neq -1 \).

Question 8. \( \int(ax^2 + bx + c)dx \)
Answer: We integrate each term separately: \( \int ax^2 dx + \int bx dx + \int c dx \). Taking out the constants, this is \( a \int x^2 dx + b \int x dx + c \int 1 dx \). Using the power rule, we get \( a \left( \frac{x^{2+1}}{2+1} \right) + b \left( \frac{x^{1+1}}{1+1} \right) + c x + C \). Simplifying the exponents and denominators yields \( \frac{ax^3}{3} + \frac{bx^2}{2} + cx + C \).
In simple words: Integrate each part of the polynomial separately. For \( ax^2 \), it becomes \( \frac{ax^3}{3} \). For \( bx \), it becomes \( \frac{bx^2}{2} \). For \( c \), it becomes \( cx \). Add C to complete the antiderivative.

Exam Tip: Remember to integrate each term of a polynomial individually. Constants multiply the integral, and 'C' is essential for indefinite integrals.

Question 9. \( \int(2x^2 + e^x) dx \)
Answer: We can separate the integral into two parts: \( \int 2x^2 dx + \int e^x dx \). Taking out the constant for the first term, we have \( 2 \int x^2 dx + \int e^x dx \). Applying the integration rules, this gives \( 2 \left( \frac{x^{2+1}}{2+1} \right) + e^x + C \). Simplifying, we obtain \( \frac{2x^3}{3} + e^x + C \).
In simple words: Integrate \( 2x^2 \) and \( e^x \) separately. For \( 2x^2 \), you get \( \frac{2x^3}{3} \). For \( e^x \), you get \( e^x \). Combine them and add C.

Exam Tip: The integral of \( e^x \) is \( e^x \) itself. Combine this with the power rule for algebraic terms.

Question 10. \( \int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx \)
Answer: First, expand the square: \( \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 = (\sqrt{x})^2 - 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) + \left(\frac{1}{\sqrt{x}}\right)^2 \). This simplifies to \( x - 2 + \frac{1}{x} \). So, the integral becomes \( \int \left( x - 2 + \frac{1}{x} \right)dx \). Now, integrate each term: \( \int x dx - \int 2 dx + \int \frac{1}{x} dx \). This results in \( \frac{x^2}{2} - 2x + \log|x| + C \).
In simple words: First, square the expression inside the integral. This will give \( x - 2 + \frac{1}{x} \). Then, integrate each term: \( x \) becomes \( \frac{x^2}{2} \), \( -2 \) becomes \( -2x \), and \( \frac{1}{x} \) becomes \( \log|x| \). Add C.

Exam Tip: Remember the square of a binomial \( (a-b)^2 = a^2 - 2ab + b^2 \). Also, the integral of \( \frac{1}{x} \) is \( \log|x| \), not \( \frac{x^0}{0} \).

Question 11. \( \int \frac{x^3 + 5x^2 - 4}{x^2} dx \)
Answer: Divide each term in the numerator by \( x^2 \): \( \int \left( \frac{x^3}{x^2} + \frac{5x^2}{x^2} - \frac{4}{x^2} \right) dx \). This simplifies to \( \int \left( x + 5 - 4x^{-2} \right) dx \). Now, integrate each term: \( \int x dx + \int 5 dx - \int 4x^{-2} dx \). Applying the power rule, we get \( \frac{x^2}{2} + 5x - 4 \left( \frac{x^{-2+1}}{-2+1} \right) + C \). Simplifying the last term: \( -4 \left( \frac{x^{-1}}{-1} \right) = 4x^{-1} = \frac{4}{x} \). So the final answer is \( \frac{x^2}{2} + 5x + \frac{4}{x} + C \).
In simple words: Separate the fraction by dividing each top part by \( x^2 \). This gives \( x + 5 - 4x^{-2} \). Then, integrate each piece. Remember \( x^{-2} \) becomes \( -x^{-1} \).

Exam Tip: For rational functions where the denominator is a single term, divide each term of the numerator by the denominator before integrating. Convert all terms to \( x^n \) form.

Question 12. \( \int \frac{x^3 + 3x + 4}{\sqrt{x}} dx \)
Answer: Rewrite the integrand by dividing each term by \( \sqrt{x} = x^{1/2} \): \( \int \left( \frac{x^3}{x^{1/2}} + \frac{3x}{x^{1/2}} + \frac{4}{x^{1/2}} \right) dx \). Using exponent rules, this becomes \( \int \left( x^{3 - 1/2} + 3x^{1 - 1/2} + 4x^{-1/2} \right) dx \), which simplifies to \( \int \left( x^{5/2} + 3x^{1/2} + 4x^{-1/2} \right) dx \). Now, integrate each term using the power rule: \( \frac{x^{5/2+1}}{5/2+1} + 3 \frac{x^{1/2+1}}{1/2+1} + 4 \frac{x^{-1/2+1}}{-1/2+1} + C \). This yields \( \frac{x^{7/2}}{7/2} + 3 \frac{x^{3/2}}{3/2} + 4 \frac{x^{1/2}}{1/2} + C \). Finally, simplify the coefficients: \( \frac{2}{7} x^{7/2} + 3 \left( \frac{2}{3} \right) x^{3/2} + 4(2) x^{1/2} + C \). The result is \( \frac{2}{7} x^{7/2} + 2x^{3/2} + 8x^{1/2} + C \).
In simple words: Divide each term on top by \( \sqrt{x} \), which is \( x^{1/2} \). Subtract the powers. Then, integrate each new term using the power rule, adding 1 to the power and dividing by the new power. Simplify the fractions.

Exam Tip: Convert square roots to fractional exponents \( (x^{1/2}) \) to apply the power rule for integration consistently. Remember to simplify the coefficients after integration.

Question 13. \( \int \frac{x^3 - x^2 + x - 1}{x - 1} dx \)
Answer: First, factor the numerator: \( x^3 - x^2 + x - 1 = x^2(x-1) + 1(x-1) = (x-1)(x^2 + 1) \). So the integral becomes \( \int \frac{(x-1)(x^2 + 1)}{x-1} dx \). We can cancel the \( (x-1) \) term (assuming \( x \neq 1 \)): \( \int (x^2 + 1) dx \). Now, integrate each term: \( \int x^2 dx + \int 1 dx \). Applying the power rule, this gives \( \frac{x^{2+1}}{2+1} + x + C \). Simplifying, we get \( \frac{x^3}{3} + x + C \).
In simple words: Factor the top part of the fraction. You will see that \( (x-1) \) is a common factor. Cancel \( (x-1) \) from the top and bottom. Then, integrate the remaining expression, \( (x^2 + 1) \), term by term, and add C.

Exam Tip: Always try to simplify the integrand by factoring or algebraic manipulation, especially for rational functions. This can greatly simplify the integration process.

Question 14. \( \int (1-x)\sqrt{x} dx \)
Answer: First, distribute \( \sqrt{x} = x^{1/2} \) into the parenthesis: \( \int (1-x)x^{1/2} dx = \int (x^{1/2} - x \cdot x^{1/2}) dx \). This simplifies to \( \int (x^{1/2} - x^{1+1/2}) dx = \int (x^{1/2} - x^{3/2}) dx \). Now, integrate each term using the power rule: \( \frac{x^{1/2+1}}{1/2+1} - \frac{x^{3/2+1}}{3/2+1} + C \). This results in \( \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} + C \). Finally, rewrite with the coefficients: \( \frac{2}{3} x^{3/2} - \frac{2}{5} x^{5/2} + C \).
In simple words: Multiply \( \sqrt{x} \) into the bracket, writing \( \sqrt{x} \) as \( x^{1/2} \). Combine the powers of \( x \). Then, integrate each term using the power rule: add 1 to the power and divide by the new power.

Exam Tip: Always convert square roots and other radicals to fractional exponents before performing integration. This makes applying the power rule straightforward.

Question 15. \( \int \sqrt{x}(3x^2 + 2x + 3)dx \)
Answer: First, distribute \( \sqrt{x} = x^{1/2} \) into the parenthesis: \( \int x^{1/2}(3x^2 + 2x + 3)dx = \int (3x^2 \cdot x^{1/2} + 2x \cdot x^{1/2} + 3x^{1/2})dx \). This simplifies to \( \int (3x^{2+1/2} + 2x^{1+1/2} + 3x^{1/2})dx = \int (3x^{5/2} + 2x^{3/2} + 3x^{1/2})dx \). Now, integrate each term: \( 3 \frac{x^{5/2+1}}{5/2+1} + 2 \frac{x^{3/2+1}}{3/2+1} + 3 \frac{x^{1/2+1}}{1/2+1} + C \). This gives \( 3 \frac{x^{7/2}}{7/2} + 2 \frac{x^{5/2}}{5/2} + 3 \frac{x^{3/2}}{3/2} + C \). Finally, simplify the coefficients: \( \frac{6}{7} x^{7/2} + \frac{4}{5} x^{5/2} + \frac{6}{3} x^{3/2} + C \). So the final answer is \( \frac{6}{7} x^{7/2} + \frac{4}{5} x^{5/2} + 2x^{3/2} + C \).
In simple words: Multiply \( x^{1/2} \) by each term inside the bracket, adding the powers. Then, integrate each new term using the power rule. For each term, add 1 to the exponent and divide by the new exponent. Simplify the result.

Exam Tip: Be careful with fractional exponents when adding them. Make sure to use a common denominator. Distribute fully before integrating.

Question 16. \( \int(2x - 3\cos x + e^x)dx \)
Answer: We can integrate each term separately: \( \int 2x dx - \int 3\cos x dx + \int e^x dx \). Taking out the constants, this becomes \( 2 \int x dx - 3 \int \cos x dx + \int e^x dx \). Applying the integration rules: \( 2 \left( \frac{x^{1+1}}{1+1} \right) - 3 (\sin x) + e^x + C \). Simplifying, we get \( 2 \left( \frac{x^2}{2} \right) - 3 \sin x + e^x + C \). The final expression is \( x^2 - 3 \sin x + e^x + C \).
In simple words: Integrate each term: \( 2x \) becomes \( x^2 \), \( -3\cos x \) becomes \( -3\sin x \), and \( e^x \) stays \( e^x \). Combine them and add C.

Exam Tip: Remember the integral of \( \cos x \) is \( \sin x \) (not \( -\sin x \)), and the integral of \( e^x \) is \( e^x \).

Question 17. \( \int(2x^2 - 3\sin x + 5\sqrt{x})dx \)
Answer: Integrate each term individually: \( \int 2x^2 dx - \int 3\sin x dx + \int 5\sqrt{x} dx \). We can write \( \sqrt{x} \) as \( x^{1/2} \). Taking out constants: \( 2 \int x^2 dx - 3 \int \sin x dx + 5 \int x^{1/2} dx \). Applying the integration rules: \( 2 \left( \frac{x^{2+1}}{2+1} \right) - 3 (-\cos x) + 5 \left( \frac{x^{1/2+1}}{1/2+1} \right) + C \). This yields \( \frac{2x^3}{3} + 3\cos x + 5 \left( \frac{x^{3/2}}{3/2} \right) + C \). Simplifying the last term gives \( \frac{2x^3}{3} + 3\cos x + 5 \left( \frac{2}{3} \right) x^{3/2} + C \). So the final answer is \( \frac{2x^3}{3} + 3\cos x + \frac{10}{3} x^{3/2} + C \).
In simple words: Integrate each part separately. \( 2x^2 \) becomes \( \frac{2x^3}{3} \). \( -3\sin x \) becomes \( 3\cos x \) (because the integral of \( \sin x \) is \( -\cos x \)). \( 5\sqrt{x} \) becomes \( \frac{10}{3} x^{3/2} \). Add C at the end.

Exam Tip: Remember that the integral of \( \sin x \) is \( -\cos x \). For \( \sqrt{x} \), convert to \( x^{1/2} \) and apply the power rule.

Question 18. \( \int \sec x (\sec x + \tan x)dx \)
Answer: First, distribute \( \sec x \) into the parenthesis: \( \int (\sec^2 x + \sec x \tan x) dx \). Now, integrate each term: \( \int \sec^2 x dx + \int \sec x \tan x dx \). We know the standard integrals: \( \int \sec^2 x dx = \tan x \) and \( \int \sec x \tan x dx = \sec x \). So, the integral is \( \tan x + \sec x + C \).
In simple words: Multiply \( \sec x \) by both terms inside the bracket. This gives \( \sec^2 x + \sec x \tan x \). Then, recall that the integral of \( \sec^2 x \) is \( \tan x \), and the integral of \( \sec x \tan x \) is \( \sec x \). Add C.

Exam Tip: Memorize the derivatives of basic trigonometric functions and their corresponding integrals. This question relies on recognizing standard integral forms.

Question 19. \( \int \frac{\sec^2 x}{\operatorname{cosec}^2 x} dx \)
Answer: Convert \( \sec^2 x \) and \( \operatorname{cosec}^2 x \) into terms of \( \sin x \) and \( \cos x \): \( \sec^2 x = \frac{1}{\cos^2 x} \) and \( \operatorname{cosec}^2 x = \frac{1}{\sin^2 x} \). So the expression becomes \( \int \frac{1/\cos^2 x}{1/\sin^2 x} dx = \int \frac{\sin^2 x}{\cos^2 x} dx \). This simplifies to \( \int \tan^2 x dx \). We use the identity \( \tan^2 x = \sec^2 x - 1 \). So the integral is \( \int (\sec^2 x - 1) dx \). Integrating each term: \( \int \sec^2 x dx - \int 1 dx \). This gives \( \tan x - x + C \).
In simple words: Change \( \sec^2 x \) to \( \frac{1}{\cos^2 x} \) and \( \operatorname{cosec}^2 x \) to \( \frac{1}{\sin^2 x} \). The fraction becomes \( \frac{\sin^2 x}{\cos^2 x} \), which is \( \tan^2 x \). Use the identity \( \tan^2 x = \sec^2 x - 1 \) and then integrate \( \sec^2 x \) (to \( \tan x \)) and \( -1 \) (to \( -x \)). Add C.

Exam Tip: When dealing with quotients of trigonometric functions, often converting to sine and cosine can help simplify the expression. Remember key trigonometric identities like \( \tan^2 x = \sec^2 x - 1 \).

Question 20. \( \int \frac{2 - 3 \sin x}{\cos^2 x} dx \)
Answer: Split the fraction into two terms: \( \int \left( \frac{2}{\cos^2 x} - \frac{3 \sin x}{\cos^2 x} \right) dx \). This can be rewritten using trigonometric identities: \( \int \left( 2 \sec^2 x - 3 \frac{\sin x}{\cos x} \frac{1}{\cos x} \right) dx \). This simplifies to \( \int (2 \sec^2 x - 3 \tan x \sec x) dx \). Now, integrate each term: \( 2 \int \sec^2 x dx - 3 \int \sec x \tan x dx \). Applying the standard integral forms, we get \( 2 (\tan x) - 3 (\sec x) + C \). So the final answer is \( 2 \tan x - 3 \sec x + C \).
In simple words: Break the fraction into two parts. Rewrite \( \frac{1}{\cos^2 x} \) as \( \sec^2 x \) and \( \frac{\sin x}{\cos^2 x} \) as \( \tan x \sec x \). Then, integrate \( \sec^2 x \) to get \( \tan x \) and \( \sec x \tan x \) to get \( \sec x \).

Exam Tip: When the denominator is a single trigonometric function squared, try splitting the fraction and using reciprocal and quotient identities to simplify the terms for integration.

Choose the correct answers in the following questions 21 and 22:

Question 21. The antiderivative of \( \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \) equals
(A) \( \frac{1}{3}x^{1/3} + 2x^{1/2} + C \)
(B) \( \frac{2}{3}x^{2/3} + \frac{1}{2}x^2 + C \)
(C) \( \frac{2}{3}x^{3/2} + 2x^{1/2} + C \)
(D) \( \frac{3}{2}x^{3/2} + \frac{1}{2}x^{1/2} + C \)
Answer: (C) \( \frac{2}{3}x^{3/2} + 2x^{1/2} + C \)
In simple words: To find the antiderivative, we integrate each term. \( \sqrt{x} \) is \( x^{1/2} \), and \( \frac{1}{\sqrt{x}} \) is \( x^{-1/2} \). Adding 1 to the power and dividing by the new power for each term gives \( \frac{2}{3}x^{3/2} + 2x^{1/2} + C \).

Exam Tip: Convert \( \sqrt{x} \) to \( x^{1/2} \) and \( \frac{1}{\sqrt{x}} \) to \( x^{-1/2} \) before applying the power rule of integration. Be careful with fractional arithmetic.

Question 22. If \( \frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4} \) such that \( f(2) = 0 \), then \( f(x) \) is
(A) \( x^4 + \frac{1}{x^3} - \frac{129}{8} \)
(B) \( x^3 + \frac{1}{x^4} + \frac{129}{8} \)
(C) \( x^4 + \frac{1}{x^3} + \frac{129}{8} \)
(D) \( x^3 + \frac{1}{x^4} - \frac{129}{8} \)
Answer: (A) \( x^4 + \frac{1}{x^3} - \frac{129}{8} \)
In simple words: First, integrate \( 4x^3 - 3x^{-4} \) to find \( f(x) \) with a constant C. Then, use the condition \( f(2)=0 \) to find the exact value of C.

Exam Tip: When given a derivative and a point on the original function, first integrate to find the general antiderivative with '+C', then use the given point to solve for the specific value of C.