GSEB Class 12 Maths Solutions Chapter 7 Integrals Exercise 7.9

Evaluate the following definite integrals.

Question 1. \( \int_{-1}^{1} (x + 1)dx \)
Answer: Evaluating the definite integral \( \int_{-1}^{1} (x+1)dx \), we first find the antiderivative, which is \( \frac{x^2}{2} + x \). We then substitute the upper limit (1) and subtract the result of substituting the lower limit (-1). This calculation gives us:
\( \left[\frac{x^2}{2} + x\right]_{-1}^{1} = \left(\frac{1^2}{2} + 1\right) - \left(\frac{(-1)^2}{2} + (-1)\right) \)
\( = \left(\frac{1}{2} + 1\right) - \left(\frac{1}{2} - 1\right) \)
\( = \frac{3}{2} - \left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2 \)
In simple words: To solve this integral, we first find the basic integral of \(x+1\), then put in the top number (1) and subtract what we get when we put in the bottom number (-1). The result of this calculation is 2.

Exam Tip: Remember to correctly apply the limits of integration, substituting the upper limit first and then subtracting the lower limit's value.

Question 2. \( \int_{2}^{3} \frac{1}{x} dx \)
Answer: To evaluate the definite integral \( \int_{2}^{3} \frac{1}{x} dx \), we identify that the antiderivative of \( \frac{1}{x} \) is \( \log |x| \). We then apply the limits of integration, calculating \( \log 3 - \log 2 \). Using logarithm properties, this simplifies to:
\( [\log |x|]_{2}^{3} = \log 3 - \log 2 \)
\( = \log \left(\frac{3}{2}\right) \)
In simple words: For this integral, we know the integral of \(1/x\) is \( \log x \). So we calculate \( \log 3 \) minus \( \log 2 \), which simplifies to \( \log(3/2) \).

Exam Tip: Always remember the property \( \log a - \log b = \log \left(\frac{a}{b}\right) \) for simplifying logarithmic expressions.

Question 3. \( \int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) dx \)
Answer: To solve the integral \( \int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) dx \), we first find the antiderivative of each term. This yields \( x^4 - \frac{5x^3}{3} + 3x^2 + 9x \). Next, we substitute the upper limit (2) into the antiderivative and subtract the value obtained by substituting the lower limit (1). Performing these calculations, we get:
\( \left[x^4 - \frac{5x^3}{3} + 3x^2 + 9x\right]_{1}^{2} \)
\( = \left(2^4 - \frac{5(2^3)}{3} + 3(2^2) + 9(2)\right) - \left(1^4 - \frac{5(1^3)}{3} + 3(1^2) + 9(1)\right) \)
\( = \left(16 - \frac{5 \cdot 8}{3} + 3 \cdot 4 + 18\right) - \left(1 - \frac{5}{3} + 3 + 9\right) \)
\( = \left(16 - \frac{40}{3} + 12 + 18\right) - \left(1 - \frac{5}{3} + 3 + 9\right) \)
\( = \left(46 - \frac{40}{3}\right) - \left(13 - \frac{5}{3}\right) \)
\( = \frac{138-40}{3} - \frac{39-5}{3} \)
\( = \frac{98}{3} - \frac{34}{3} = \frac{64}{3} \)
In simple words: First, we find the integral for each part of the expression. Then we put the top number (2) into the answer and subtract what we get when we put the bottom number (1) in. After doing all the math, the final answer is \( 64/3 \).

Exam Tip: For polynomial integrals, find the antiderivative of each term and be careful with arithmetic when substituting the limits.

Question 4. \( \int_{0}^{\frac{\pi}{4}} \sin 2x dx \)
Answer: To evaluate \( \int_{0}^{\frac{\pi}{4}} \sin 2x dx \), we first determine the antiderivative of \( \sin 2x \), which is \( -\frac{1}{2} \cos 2x \). We then apply the limits of integration. Substituting \( \frac{\pi}{4} \) and 0, we calculate:
\( \left[-\frac{1}{2} \cos 2x\right]_{0}^{\frac{\pi}{4}} \)
\( = -\frac{1}{2} \left(\cos \left(2 \cdot \frac{\pi}{4}\right) - \cos(2 \cdot 0)\right) \)
\( = -\frac{1}{2} \left(\cos \frac{\pi}{2} - \cos 0\right) \)
\( = -\frac{1}{2} (0 - 1) = \frac{1}{2} \)
In simple words: We integrate \( \sin 2x \) to get \( -\frac{1}{2} \cos 2x \). Then we put in the top limit \( \pi/4 \) and subtract what we get from the bottom limit 0. This gives us \( -\frac{1}{2} (0 - 1) \), which results in \( \frac{1}{2} \).

Exam Tip: Remember standard trigonometric values like \( \cos 0 = 1 \) and \( \cos \frac{\pi}{2} = 0 \) to avoid common calculation errors.

Question 5. \( \int_{0}^{\frac{\pi}{2}} \cos 2x dx \)
Answer: To evaluate \( \int_{0}^{\frac{\pi}{2}} \cos 2x dx \), we first find the antiderivative of \( \cos 2x \), which is \( \frac{1}{2} \sin 2x \). We then substitute the upper limit \( \frac{\pi}{2} \) and the lower limit 0. This leads to:
\( \left[\frac{1}{2} \sin 2x\right]_{0}^{\frac{\pi}{2}} \)
\( = \frac{1}{2} \left(\sin \left(2 \cdot \frac{\pi}{2}\right) - \sin(2 \cdot 0)\right) \)
\( = \frac{1}{2} \left(\sin \pi - \sin 0\right) \)
\( = \frac{1}{2} (0 - 0) = 0 \)
In simple words: We integrate \( \cos 2x \) to get \( \frac{1}{2} \sin 2x \). Then we put in the top limit \( \pi/2 \) and subtract what we get from the bottom limit 0. Since \( \sin \pi \) and \( \sin 0 \) are both zero, the final answer is 0.

Exam Tip: Pay close attention to the sine values at \( \pi \) and 0, as they are both zero and can simplify calculations quickly.

Question 6. \( \int_{4}^{5} e^x dx \)
Answer: To evaluate the definite integral \( \int_{4}^{5} e^x dx \), we note that the antiderivative of \( e^x \) is \( e^x \) itself. We then substitute the upper limit (5) and subtract the result of substituting the lower limit (4). This gives us:
\( [e^x]_{4}^{5} = e^5 - e^4 \)
\( = e^4(e - 1) \)
In simple words: The integral of \( e^x \) is just \( e^x \). So we calculate \( e^5 \) minus \( e^4 \), and we can write this as \( e^4(e-1) \).

Exam Tip: Always remember that the derivative and integral of \( e^x \) are both \( e^x \), simplifying calculations significantly.

Question 7. \( \int_{0}^{\frac{\pi}{4}} \tan x dx \)
Answer: To evaluate \( \int_{0}^{\frac{\pi}{4}} \tan x dx \), we find the antiderivative of \( \tan x \), which is \( -\log |\cos x| \). We then substitute the limits \( \frac{\pi}{4} \) and 0. This calculation leads to:
\( [-\log |\cos x|]_{0}^{\frac{\pi}{4}} = -\left(\log \left|\cos \frac{\pi}{4}\right| - \log |\cos 0|\right) \)
\( = -\left(\log \left|\frac{1}{\sqrt{2}}\right| - \log |1|\right) \)
\( = -\left(\log \left(\frac{1}{\sqrt{2}}\right) - 0\right) \)
\( = -\log \left(2^{-\frac{1}{2}}\right) = \frac{1}{2} \log 2 \)
In simple words: We integrate \( \tan x \) to get \( -\log |\cos x| \). Then we put in the top limit \( \pi/4 \) and subtract what we get from the bottom limit 0. This simplifies to \( \frac{1}{2} \log 2 \).

Exam Tip: Recall the integral of \( \tan x \) and simplify logarithmic expressions using properties like \( \log (a^b) = b \log a \).

Question 8. \( \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x dx \)
Answer: To evaluate the definite integral \( \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x dx \), we first find the antiderivative of \( \operatorname{cosec} x \), which is \( \log |\operatorname{cosec} x - \cot x| \). We then apply the limits of integration \( \frac{\pi}{4} \) and \( \frac{\pi}{6} \). This involves calculating:
\( [\log |\operatorname{cosec} x - \cot x|]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \)
\( = \log \left|\operatorname{cosec} \frac{\pi}{4} - \cot \frac{\pi}{4}\right| - \log \left|\operatorname{cosec} \frac{\pi}{6} - \cot \frac{\pi}{6}\right| \)
\( = \log |\sqrt{2} - 1| - \log |2 - \sqrt{3}| \)
\( = \log \left|\frac{\sqrt{2} - 1}{2 - \sqrt{3}}\right| \)
In simple words: We integrate \( \operatorname{cosec} x \) to get \( \log |\operatorname{cosec} x - \cot x| \). Then we put in the top limit \( \pi/4 \) and subtract what we get from the bottom limit \( \pi/6 \). After plugging in the values, we simplify it using log rules to get \( \log \left|\frac{\sqrt{2} - 1}{2 - \sqrt{3}}\right| \).

Exam Tip: Be sure to memorize the standard integral formulas for trigonometric functions and know common trigonometric values at special angles.

Question 9. \( \int_{0}^{1} \frac{dx}{\sqrt{1-x^{2}}} \)
Answer: To evaluate \( \int_{0}^{1} \frac{dx}{\sqrt{1-x^{2}}} \), we recognize that the integral of \( \frac{1}{\sqrt{1-x^2}} \) is \( \sin^{-1} x \). Applying the limits of integration, we compute:
\( [\sin^{-1} x]_{0}^{1} = \sin^{-1} 1 - \sin^{-1} 0 \)
\( = \frac{\pi}{2} - 0 = \frac{\pi}{2} \)
In simple words: We know the integral of \( 1/\sqrt{1-x^2} \) is \( \sin^{-1} x \). So, we calculate \( \sin^{-1} 1 \) minus \( \sin^{-1} 0 \), which gives us \( \pi/2 \) minus 0, making the final answer \( \pi/2 \).

Exam Tip: Recognize standard inverse trigonometric integral forms and their respective derivatives, especially \( \int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1} \left(\frac{x}{a}\right) \).

Question 10. \( \int_{0}^{1} \frac{dx}{1+x^{2}} \)
Answer: To evaluate \( \int_{0}^{1} \frac{dx}{1+x^{2}} \), we recognize that the antiderivative of \( \frac{1}{1+x^2} \) is \( \tan^{-1} x \). Applying the limits of integration, we compute:
\( [\tan^{-1} x]_{0}^{1} = \tan^{-1} 1 - \tan^{-1} 0 \)
\( = \frac{\pi}{4} - 0 = \frac{\pi}{4} \)
In simple words: We know the integral of \( 1/(1+x^2) \) is \( \tan^{-1} x \). So, we calculate \( \tan^{-1} 1 \) minus \( \tan^{-1} 0 \), which gives us \( \pi/4 \) minus 0, making the final answer \( \pi/4 \).

Exam Tip: Remember the integral of \( \frac{1}{1+x^2} \) as it is a common inverse trigonometric integral, leading to \( \tan^{-1} x \).

Question 11. \( \int_{2}^{3} \frac{dx}{x^{2}-1} \)
Answer: To evaluate \( \int_{2}^{3} \frac{dx}{x^{2}-1} \), we use the standard integral formula for \( \frac{1}{x^2-a^2} \). The antiderivative is \( \frac{1}{2} \log \left|\frac{x-1}{x+1}\right| \). We then substitute the limits 3 and 2. This leads to:
\( \left[\frac{1}{2} \log \left|\frac{x-1}{x+1}\right|\right]_{2}^{3} \)
\( = \frac{1}{2} \left(\log \left|\frac{3-1}{3+1}\right| - \log \left|\frac{2-1}{2+1}\right|\right) \)
\( = \frac{1}{2} \left(\log \left|\frac{2}{4}\right| - \log \left|\frac{1}{3}\right|\right) \)
\( = \frac{1}{2} \left(\log \left(\frac{1}{2}\right) - \log \left(\frac{1}{3}\right)\right) \)
\( = \frac{1}{2} \log \left(\frac{1/2}{1/3}\right) = \frac{1}{2} \log \left(\frac{1}{2} \cdot 3\right) = \frac{1}{2} \log \left(\frac{3}{2}\right) \)
In simple words: We use a special formula to integrate \( 1/(x^2-1) \), which gives us \( \frac{1}{2} \log \left|\frac{x-1}{x+1}\right| \). Then we plug in the top limit (3) and subtract what we get from the bottom limit (2). After simplifying, the answer becomes \( \frac{1}{2} \log \left(\frac{3}{2}\right) \).

Exam Tip: Know the partial fraction decomposition formula for \( \frac{1}{x^2-a^2} \) which leads to logarithmic terms.

Question 12. \( \int_{0}^{\frac{\pi}{2}} \cos^2 x dx \)
Answer: To evaluate \( \int_{0}^{\frac{\pi}{2}} \cos^2 x dx \), we first use the trigonometric identity \( \cos^2 x = \frac{1+\cos 2x}{2} \). The integral then becomes:
\( \int_{0}^{\frac{\pi}{2}} \frac{1+\cos 2x}{2} dx \quad [\because \cos 2x = 2 \cos^2 x - 1] \)
\( = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1+\cos 2x) dx \)
Integrating this, we get:
\( = \frac{1}{2} \left[x + \frac{\sin 2x}{2}\right]_{0}^{\frac{\pi}{2}} \)
Substituting the limits, we compute:
\( = \frac{1}{2} \left(\left(\frac{\pi}{2} + \frac{\sin \left(2 \cdot \frac{\pi}{2}\right)}{2}\right) - \left(0 + \frac{\sin(2 \cdot 0)}{2}\right)\right) \)
\( = \frac{1}{2} \left(\left(\frac{\pi}{2} + \frac{\sin \pi}{2}\right) - \left(0 + \frac{\sin 0}{2}\right)\right) \)
\( = \frac{1}{2} \left(\left(\frac{\pi}{2} + 0\right) - (0 + 0)\right) = \frac{1}{2} \left(\frac{\pi}{2}\right) = \frac{\pi}{4} \)
In simple words: We change \( \cos^2 x \) to \( (1+\cos 2x)/2 \) using a trig rule. Then we integrate this new expression. We put in the top limit \( \pi/2 \) and subtract what we get from the bottom limit 0. After calculating and simplifying, the answer is \( \pi/4 \).

Exam Tip: When integrating even powers of sine or cosine, use power-reducing identities like \( \cos^2 x = \frac{1+\cos 2x}{2} \).

Question 13. \( \int_{2}^{3} \frac{x dx}{x^{2}+1} \)
Answer: To evaluate \( \int_{2}^{3} \frac{x dx}{x^{2}+1} \), we use a substitution method. Let \( t = x^2+1 \), which implies that \( dt = 2x dx \). Consequently, \( x dx = \frac{1}{2} dt \). We also need to change the limits of integration:
When \( x=2, t = 2^2+1 = 5 \).
When \( x=3, t = 3^2+1 = 10 \).
The integral then transforms to:
\( I = \int_{5}^{10} \frac{1}{t} \frac{1}{2} dt = \frac{1}{2} [\log |t|]_{5}^{10} \)
Applying the new limits, this becomes:
\( = \frac{1}{2} (\log 10 - \log 5) \)
\( = \frac{1}{2} \log \left(\frac{10}{5}\right) = \frac{1}{2} \log 2 \)
In simple words: We use substitution here. Let \( t \) be \( x^2+1 \), so \( dx \) changes too. We also change the top and bottom numbers (limits) for \( t \). The integral then becomes simpler. After integrating and putting in the new limits, we get \( \frac{1}{2} (\log 10 - \log 5) \), which is \( \frac{1}{2} \log 2 \).

Exam Tip: Always remember to change the limits of integration when performing a substitution in definite integrals.

Question 14. \( \int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} dx \)
Answer: To evaluate \( \int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} dx \), we first split the integral into two parts:
\( \int_{0}^{1} \frac{2x+3}{5x^2+1} dx = \int_{0}^{1} \left(\frac{2x}{5x^2+1} + \frac{3}{5x^2+1}\right) dx \)
\( = \int_{0}^{1} \frac{2x}{5x^2+1} dx + 3 \int_{0}^{1} \frac{1}{5x^2+1} dx \)
For the first integral, we use a substitution \( u = 5x^2+1 \), which leads to:
\( = \frac{1}{5} \int_{0}^{1} \frac{10x}{5x^2+1} dx = \frac{1}{5} [\log |5x^2+1|]_{0}^{1} \)
For the second part, we factor out 5 from the denominator to get the form \( \int \frac{1}{x^2+a^2} dx \), which integrates to \( \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) \):
\( = \frac{3}{5} \int_{0}^{1} \frac{1}{x^2+\left(\frac{1}{\sqrt{5}}\right)^2} dx = \frac{3}{5} \left[\frac{1}{\frac{1}{\sqrt{5}}} \tan^{-1} \left(\frac{x}{\frac{1}{\sqrt{5}}}\right)\right]_{0}^{1} \)
Combining and applying the limits to both parts:
\( = \frac{1}{5} (\log |5(1)^2+1| - \log |5(0)^2+1|) + \frac{3}{5} \sqrt{5} [\tan^{-1} (\sqrt{5}x)]_{0}^{1} \)
\( = \frac{1}{5} (\log 6 - \log 1) + \frac{3}{\sqrt{5}} (\tan^{-1} \sqrt{5} - \tan^{-1} 0) \)
\( = \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1} \sqrt{5} \)
In simple words: We break this integral into two parts. The first part uses a substitution to get a logarithm. The second part uses the inverse tangent formula after some rewriting. Then we apply the limits (0 and 1) to both parts and add them up to get \( \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1} \sqrt{5} \).

Exam Tip: When the numerator is a linear function and the denominator is a quadratic with no real roots, split the integral into two parts: one for a logarithmic term and one for an inverse tangent term.

Question 15. \( \int_{0}^{1} x e^{x^2} dx \)
Answer: To evaluate \( \int_{0}^{1} x e^{x^2} dx \), we employ a substitution method. Let \( t = x^2 \), which means \( dt = 2x dx \). Consequently, \( x dx = \frac{1}{2} dt \). We also adjust the limits of integration:
When \( x=0, t = 0^2 = 0 \).
When \( x=1, t = 1^2 = 1 \).
The integral thus becomes:
\( I = \int_{0}^{1} e^t \frac{1}{2} dt = \frac{1}{2} \int_{0}^{1} e^t dt \)
Integrating \( e^t \) yields \( e^t \), so we calculate:
\( = \frac{1}{2} [e^t]_{0}^{1} = \frac{1}{2} (e^1 - e^0) \)
\( = \frac{1}{2} (e - 1) \)
In simple words: We use substitution here. Let \( t \) be \( x^2 \), so \( dx \) changes, and the limits of integration (0 and 1) stay the same. The integral becomes \( \frac{1}{2} \int_{0}^{1} e^t dt \). After integrating, we get \( \frac{1}{2} (e^1 - e^0) \), which simplifies to \( \frac{1}{2} (e - 1) \).

Exam Tip: Look for opportunities to use u-substitution, especially when one part of the integrand is the derivative of another part.

Question 16. \( \int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} dx \)
Answer: To evaluate \( \int_{1}^{2} \frac{5x^2}{x^2+4x+3} dx \), we first perform polynomial division to simplify the integrand into \( 5 - \frac{20x+15}{x^2+4x+3} \). Then, we use partial fraction decomposition for \( \frac{20x+15}{x^2+4x+3} \). We can factor the denominator as \( (x+1)(x+3) \).
Let \( \frac{20x+15}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3} \).
Multiplying by \( (x+1)(x+3) \): \( 20x+15 = A(x+3) + B(x+1) \).
Putting \( x=-1 \): \( 20(-1)+15 = A(-1+3) \implies -5 = 2A \implies A = -\frac{5}{2} \).
Putting \( x=-3 \): \( 20(-3)+15 = B(-3+1) \implies -45 = -2B \implies B = \frac{45}{2} \).
So, \( \frac{20x+15}{x^2+4x+3} = -\frac{5}{2(x+1)} + \frac{45}{2(x+3)} \).
Therefore, the integral becomes:
\( I = \int_{1}^{2} \left(5 - \left(-\frac{5}{2(x+1)} + \frac{45}{2(x+3)}\right)\right) dx \)
\( I = \int_{1}^{2} \left(5 + \frac{5}{2(x+1)} - \frac{45}{2(x+3)}\right) dx \)
Integrating each term and applying the limits:
\( = \left[5x + \frac{5}{2} \log|x+1| - \frac{45}{2} \log|x+3|\right]_{1}^{2} \)
\( = \left(5(2) + \frac{5}{2} \log(2+1) - \frac{45}{2} \log(2+3)\right) - \left(5(1) + \frac{5}{2} \log(1+1) - \frac{45}{2} \log(1+3)\right) \)
\( = \left(10 + \frac{5}{2} \log 3 - \frac{45}{2} \log 5\right) - \left(5 + \frac{5}{2} \log 2 - \frac{45}{2} \log 4\right) \)
\( = 5 + \frac{5}{2} (\log 3 - \log 2) - \frac{45}{2} (\log 5 - \log 4) \)
\( = 5 + \frac{5}{2} \log \left(\frac{3}{2}\right) - \frac{45}{2} \log \left(\frac{5}{4}\right) \)
In simple words: First, we divide the polynomials in the integral. Then, we use partial fractions to break down the remaining fraction into simpler terms. We find the values for A and B. After putting everything back into the integral, we integrate each part separately. Finally, we apply the limits of 1 and 2 and simplify the logarithmic expressions to get the final answer.

Exam Tip: For rational functions where the degree of the numerator is greater than or equal to the degree of the denominator, always perform polynomial long division first before applying partial fractions.

Question 17. \( \int_{0}^{\frac{\pi}{4}} (2\sec^2 x + x^3 + 2) dx \)
Answer: To evaluate \( \int_{0}^{\frac{\pi}{4}} (2\sec^2 x + x^3 + 2) dx \), we integrate each term separately. The integral of \( 2\sec^2 x \) is \( 2\tan x \), for \( x^3 \) it is \( \frac{x^4}{4} \), and for 2 it is \( 2x \). We then apply the limits of integration, \( \frac{\pi}{4} \) and 0. Substituting these values, we calculate:
\( \left[2\tan x + \frac{x^4}{4} + 2x\right]_{0}^{\frac{\pi}{4}} \)
\( = \left(2\tan \frac{\pi}{4} + \frac{(\frac{\pi}{4})^4}{4} + 2\left(\frac{\pi}{4}\right)\right) - \left(2\tan 0 + \frac{0^4}{4} + 2(0)\right) \)
\( = \left(2(1) + \frac{\pi^4}{256 \cdot 4} + \frac{\pi}{2}\right) - (0+0+0) \)
\( = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2} \)
In simple words: We integrate each part of the expression: \( 2\sec^2 x \), \( x^3 \), and 2. Then, we put in the top limit \( \pi/4 \) and subtract what we get from the bottom limit 0. After doing all the math, the final answer is \( 2 + \frac{\pi^4}{1024} + \frac{\pi}{2} \).

Exam Tip: When integrating a sum of functions, integrate each term individually and then combine the results.

Question 18. \( \int_{0}^{\pi} \left(\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}\right) dx \)
Answer: To evaluate \( \int_{0}^{\pi} \left(\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}\right) dx \), we first use the trigonometric identity \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \). This transforms the integrand into \( -\cos x \). So, the integral becomes:
\( \int_{0}^{\pi} -\cos x dx \)
Integrating \( -\cos x \) yields \( -\sin x \). Applying the limits, we compute:
\( -[\sin x]_{0}^{\pi} = -(\sin \pi - \sin 0) \)
\( = -(0 - 0) = 0 \)
In simple words: We change the expression inside the integral using a trig rule (specifically, \( \cos x = \cos^2(x/2) - \sin^2(x/2) \)). This makes the integral \( \int_{0}^{\pi} -\cos x dx \). Then we integrate \( -\cos x \) to get \( -\sin x \). When we put in the limits \( \pi \) and 0, both \( \sin \pi \) and \( \sin 0 \) are zero, so the final answer is 0.

Exam Tip: Look for opportunities to simplify trigonometric integrands using identities before integrating.

Question 19. \( \int_{0}^{2} \frac{6 x+3}{x^{2}+4} dx \)
Answer: To evaluate \( \int_{0}^{2} \frac{6 x+3}{x^{2}+4} dx \), we separate the integrand into two fractions:
\( \int_{0}^{2} \frac{6x+3}{x^2+4} dx = \int_{0}^{2} \frac{6x}{x^2+4} dx + \int_{0}^{2} \frac{3}{x^2+4} dx \)
For the first integral, we adjust the numerator to be \( 3 \cdot 2x \), which is suitable for a logarithm integration, giving:
\( = 3 \int_{0}^{2} \frac{2x}{x^2+4} dx = 3 [\log(x^2+4)]_{0}^{2} \)
For the second integral, we use the \( \tan^{-1} \) formula for \( \int \frac{1}{x^2+a^2} dx \), yielding:
\( = 3 \int_{0}^{2} \frac{1}{x^2+2^2} dx = 3 \left[\frac{1}{2} \tan^{-1} \left(\frac{x}{2}\right)\right]_{0}^{2} \)
Applying the limits to both parts and simplifying:
\( = 3 (\log(2^2+4) - \log(0^2+4)) + \frac{3}{2} \left(\tan^{-1} \left(\frac{2}{2}\right) - \tan^{-1} \left(\frac{0}{2}\right)\right) \)
\( = 3 (\log 8 - \log 4) + \frac{3}{2} (\tan^{-1} 1 - \tan^{-1} 0) \)
\( = 3 \log \left(\frac{8}{4}\right) + \frac{3}{2} \left(\frac{\pi}{4} - 0\right) \)
\( = 3 \log 2 + \frac{3\pi}{8} \)
In simple words: We split the integral into two parts. The first part integrates to \( 3 \log(x^2+4) \). The second part uses the inverse tangent formula. Then we plug in the limits (0 and 2) for both parts. After some calculations and using log properties, we get \( 3 \log 2 + \frac{3\pi}{8} \).

Exam Tip: When the denominator is \( x^2+a^2 \) and the numerator has both \( x \) and a constant term, split the fraction to handle the \( x \) term via substitution (logarithm) and the constant term via inverse tangent.

Question 20. \( \int_{0}^{1} \left(x e^x + \sin \frac{\pi x}{2}\right) dx \)
Answer: To evaluate \( \int_{0}^{1} \left(x e^x + \sin \frac{\pi x}{2}\right) dx \), we split it into two separate integrals:
\( I = \int_{0}^{1} x e^x dx + \int_{0}^{1} \sin \frac{\pi x}{2} dx \).
For \( \int_{0}^{1} x e^x dx \): Use integration by parts, with \( u=x, dv=e^x dx \implies du=dx, v=e^x \).
\( [xe^x]_{0}^{1} - \int_{0}^{1} e^x dx = [xe^x]_{0}^{1} - [e^x]_{0}^{1} \)
\( = (1 \cdot e^1 - 0 \cdot e^0) - (e^1 - e^0) \)
\( = (e - 0) - (e - 1) = e - e + 1 = 1 \)
For \( \int_{0}^{1} \sin \frac{\pi x}{2} dx \):
\( = \left[-\frac{\cos \frac{\pi x}{2}}{\frac{\pi}{2}}\right]_{0}^{1} = -\frac{2}{\pi} \left[\cos \frac{\pi x}{2}\right]_{0}^{1} \)
\( = -\frac{2}{\pi} \left(\cos \frac{\pi}{2} - \cos 0\right) \)
\( = -\frac{2}{\pi} (0 - 1) = \frac{2}{\pi} \)
Combining results:
\( I = 1 + \frac{2}{\pi} \)
In simple words: We break this integral into two parts. The first part, \( \int x e^x dx \), is solved by integration by parts, giving 1. The second part, \( \int \sin(\pi x/2) dx \), is integrated and gives \( 2/\pi \). Adding these together, the total answer is \( 1 + 2/\pi \).

Exam Tip: Be prepared to use different integration techniques, like integration by parts and substitution, within the same problem for different terms.

Question 21. \( \int_{1}^{\sqrt{3}} \frac{dx}{1+x^{2}} \) equals
(a) \( \frac{\pi}{3} \)
(b) \( \frac{2\pi}{3} \)
(c) \( \frac{\pi}{6} \)
(d) \( \frac{\pi}{12} \)
Answer: (d) \( \frac{\pi}{12} \)
To evaluate \( \int_{1}^{\sqrt{3}} \frac{dx}{1+x^{2}} \), we know that the integral of \( \frac{1}{1+x^2} \) is \( \tan^{-1} x \). Applying the limits \( \sqrt{3} \) and 1, we calculate:
\( [\tan^{-1} x]_{1}^{\sqrt{3}} = \tan^{-1} \sqrt{3} - \tan^{-1} 1 \)
\( = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \)
Therefore, option (d) is the correct choice.
In simple words: The integral of \( 1/(1+x^2) \) is \( \tan^{-1} x \). We put in the top limit \( \sqrt{3} \) and subtract what we get from the bottom limit 1. This gives us \( \pi/3 - \pi/4 \), which equals \( \pi/12 \). So, (d) is the right answer.

Exam Tip: Knowing the common values of inverse trigonometric functions for angles like \( \frac{\pi}{3} \) and \( \frac{\pi}{4} \) is crucial for quickly solving such MCQs.

Question 22. \( \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}} \) equals
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{12} \)
(c) \( \frac{\pi}{24} \)
(d) \( \frac{\pi}{4} \)
Answer: (c) \( \frac{\pi}{24} \)
To evaluate \( \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}} \), we first rewrite the denominator as \( 9(\frac{4}{9}+x^2) \) to match the form \( \frac{1}{a^2+x^2} \). The integral becomes:
\( \int_{0}^{\frac{2}{3}} \frac{dx}{4+9x^{2}} = \int_{0}^{\frac{2}{3}} \frac{dx}{9\left(\frac{4}{9}+x^{2}\right)} \)
\( = \frac{1}{9} \int_{0}^{\frac{2}{3}} \frac{dx}{\left(\frac{2}{3}\right)^2+x^{2}} \)
Applying the formula for \( \tan^{-1} \), with \( a = \frac{2}{3} \):
\( = \frac{1}{9} \left[\frac{1}{\frac{2}{3}} \tan^{-1} \left(\frac{x}{\frac{2}{3}}\right)\right]_{0}^{\frac{2}{3}} \)
\( = \frac{1}{9} \cdot \frac{3}{2} \left[\tan^{-1} \left(\frac{3x}{2}\right)\right]_{0}^{\frac{2}{3}} \)
\( = \frac{1}{6} \left(\tan^{-1} \left(\frac{3 \cdot \frac{2}{3}}{2}\right) - \tan^{-1} \left(\frac{3 \cdot 0}{2}\right)\right) \)
\( = \frac{1}{6} (\tan^{-1} 1 - \tan^{-1} 0) \)
\( = \frac{1}{6} \left(\frac{\pi}{4} - 0\right) = \frac{\pi}{24} \)
Therefore, option (c) is the correct choice.
In simple words: First, we rewrite the integral to fit the inverse tangent formula by factoring out 9 from the denominator. This gives us \( \frac{1}{6} \left[\tan^{-1} \left(\frac{3x}{2}\right)\right]_{0}^{\frac{2}{3}} \). Then we put in the top limit \( 2/3 \) and subtract what we get from the bottom limit 0. This results in \( \frac{1}{6} (\pi/4 - 0) \), which is \( \pi/24 \). So, (c) is the right answer.

Exam Tip: Always transform the denominator into the standard form \( a^2+x^2 \) by factoring out coefficients before applying the inverse tangent integral formula.