Exercise 7.8 (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths
Updated On 26-08-2025 By Lithanya
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NCERT Class 12 Maths Solutions: Chapter 7 Integrals | Free PDF Download
Ex 7.8 Question 1.
$\int_{-1}^1(x+1) d x$
Answer.
$\int_{-1}^1(x+1) d x$
$
\begin{aligned}
& =\left(\frac{x^2}{2}+x\right)_{-1}^1 \\
& =\left(\frac{1^2}{2}+1\right)-\left(\frac{(-1)^2}{2}-1\right) \\
& =\frac{1}{2}+1-\left(\frac{1}{2}-1\right) \\
& =\frac{1}{2}+1-\frac{1}{2}+1
\end{aligned}
$
$=2$ Ans.
Ex 7.8 Question 2.
$\int_2^3 \frac{1}{x} d x$
Answer.
$\int_2^3 \frac{1}{x} d x$
$\begin{aligned}
& =(\log |x|)_2^3 \\
& =\log |3|-\log |2| \\
& =\log 3-\log 2 \\
& =\log \frac{3}{2} \text { Ans. }
\end{aligned}$
Ex 7.8 Question 3.
$\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x$
$
\begin{aligned}
& \text { Answer. } \int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x \\
& =\left(4 \frac{x^4}{4}-5 \frac{x^3}{3}+6 \frac{x^2}{2}+9 x\right)_1^2 \\
& =\left(x^4-\frac{5}{3} x^3+3 x^2+9 x\right)_1^2 \\
& =\left(2^4-\frac{5}{3}(2)^3+3(2)^2+9(2)\right)-\left(1-\frac{5}{3}+3+9\right) \\
& =\left(16-\frac{40}{3}+12+18\right)-\left(13-\frac{5}{3}\right) \\
& =46-\frac{40}{3}-\left(13-\frac{5}{3}\right) \\
& =46-\frac{40}{3}-13+\frac{5}{3} \\
& =33-\frac{40}{3}+\frac{5}{3}
\end{aligned}
$
$\begin{aligned}
& =\frac{99-40+5}{3} \\
& =\frac{64}{3} \text { Ans. }
\end{aligned}$
Ex 7.8 Question 4.
$\int_0^{\frac{\pi}{4}} \sin 2 x d x$
Answer.
$\int_0^{\frac{\pi}{4}} \sin 2 x d x$
$
\begin{aligned}
& =\left(\frac{-\cos 2 x}{2}\right)_0^{\frac{\pi}{4}} \\
& =\frac{-\cos \frac{\pi}{2}}{2}-\left(\frac{-\cos 0^{\circ}}{2}\right) \\
& =\frac{0}{2}-\left(\frac{-1}{2}\right) \\
& =0+\frac{1}{2} \\
& =\frac{1}{2} \text { Ans. }
\end{aligned}
$
Ex 7.8 Question 5.
$\int_0^{\frac{\pi}{2}} \cos 2 x d x$
Answer.
$\int_0^{\frac{\pi}{2}} \cos 2 x d x$
$
\begin{aligned}
& =\left(\frac{\sin 2 x}{2}\right)_0^{\frac{\pi}{2}} \\
& =\frac{\sin \pi}{2}-\frac{\sin 0^{\circ}}{2} \\
& =\frac{0}{2}-\frac{0}{2} \\
& =0 \text { Ans. }
\end{aligned}
$
Ex 7.8 Question 6.
$\int_4^5 e^x d x$
Answer.
$\int_4^5 e^x d x$
$
\begin{aligned}
& =\left(e^x\right)_4^5 \\
& =e^5-e^4 \\
& =e^4(e-1) \text { Ans. }
\end{aligned}
$
Ex 7.8 Question 7.
$\int_0^{\frac{\pi}{4}} \tan x d x$
Answer.
$\int_0^{\frac{\pi}{4}} \tan x d x$
$\begin{aligned}
& =(\log |\sec x|)_0^{\frac{\pi}{4}} \\
& =\log \left|\sec \frac{\pi}{4}\right|-\log \left|\sec 0^{\circ}\right| \\
& =\log |\sqrt{2}|-\log |1| \\
& =\log \sqrt{2}-\log 1 \\
& =\log 2^{\frac{1}{2}}-0 \\
& =\frac{1}{2} \log 2 \text { Ans. }
\end{aligned}$
Ex 7.8 Question 8.
$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x$
Answer.
$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x$
$
\begin{aligned}
& =(\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 (\sqrt{2}-1)-\log (2-\sqrt{3})
\end{aligned}
$
$=\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right) \text { Ans. }$
Ex 7.8 Question 9
$\begin{aligned}
& \text {} \int_0^1 \frac{d x}{\sqrt{1-x^2}} \\
& \text { Answer. } \int_0^1 \frac{d x}{\sqrt{1-x^2}} \\
& =\left(\sin ^{-1} x\right)_0^1 \\
& =\sin ^{-1} 1-\sin ^{-1} 0 \\
& =\frac{\pi}{2}-0 \\
& =\frac{\pi}{2} \text { Ans. }
\end{aligned}$
Ex 7.8 Question 10
$\int_0^1 \frac{d x}{1+x^2}$
Answer.
$\int_0^1 \frac{d x}{1+x^2}$
$
\begin{aligned}
& =\left(\tan ^{-1} x\right)_0^1 \\
& =\tan ^{-1} 1-\tan ^{-1} 0 \\
& =\frac{\pi}{4}-0 \\
& =\frac{\pi}{4} \text { Ans. }
\end{aligned}
$
Ex 7.8 Question 11.
$\int_2^3 \frac{d x}{x^2-1}$
Answer.
$\int_2^3 \frac{d x}{x^2-1}=\int_2^3 \frac{1}{x^2-1^2} d x$
$
\begin{aligned}
& =\left(\frac{1}{2(1)} \log \left|\frac{x-1}{x+1}\right|\right)_2^3 \\
& =\frac{1}{2} \log \left|\frac{3-1}{3+1}\right|-\frac{1}{2} \log \left|\frac{2-1}{2+1}\right| \\
& =\frac{1}{2} \log \left|\frac{1}{2}\right|-\frac{1}{2} \log \left|\frac{1}{3}\right| \\
& =\frac{1}{2}\left(\log \frac{1}{2}-\log \frac{1}{3}\right) \\
& =\frac{1}{2} \log \frac{1 / 2}{1 / 3} \\
& =\frac{1}{2} \log \frac{3}{2} \text { Ans. }
\end{aligned}
$
Ex 7.8 Question 12.
$\int_0^{\frac{\pi}{2}} \cos ^2 c d x$
Answer.
$\int_0^{\frac{\pi}{2}} \cos ^2 x d x$
$\begin{aligned}
& =\int_0^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x \\
& =\frac{1}{2} \int_0^{\frac{\pi}{2}}(1+\cos 2 x) d x \\
& =\frac{1}{2}\left(x+\frac{\sin 2 x}{2}\right)_0^{\frac{\pi}{2}} \\
& =\frac{1}{2}\left[\frac{\pi}{2}+\frac{1}{2} \sin \pi-\left(0+\frac{1}{2} \sin 0^{\circ}\right)\right] \\
& =\frac{1}{2}\left[\frac{\pi}{2}+0-0\right] \\
& =\frac{\pi}{4} \text { Ans. }
\end{aligned}$
Ex 7.8 Question 13.
$\int_2^3 \frac{x d x}{x^2+1}$
Answer.
$\int_2^3 \frac{x d x}{x^2+1}=\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1} d x$
$
\begin{aligned}
& =\frac{1}{2}\left(\log \left|x^2+1\right|\right)_2^3 \\
& =\frac{1}{2}(\log |10|-\log |5|) \\
& =\frac{1}{2}(\log 10-\log 5)
\end{aligned}
$
$\begin{aligned}
& =\frac{1}{2} \log \frac{10}{5} \\
& =\frac{1}{2} \log 2 \text { Ans. }
\end{aligned}$
Ex 7.8 Question 14.
$\int_0^1 \frac{2 x+3}{5 x^2+1} d x$
Answer.
$\int_0^1 \frac{2 x+3}{5 x^2+1} d x$
$
\begin{aligned}
& =\int_0^1\left(\frac{2 x}{5 x^2+1}+\frac{3}{5 x^2+1}\right) d x \\
& =\int_0^1 \frac{2 x}{5 x^2+1} d x+\int_0^1 \frac{3}{5 x^2+1} d x \\
& =\int_0^1 \frac{2 x}{5 x^2+1} d x+3 \int_0^1 \frac{1}{(\sqrt{5} x)^2+1^2} d x \\
& =\frac{1}{5} \int_0^1 \frac{10 x}{5 x^2+1} d x+3 \int_0^1 \frac{1}{(\sqrt{5} x)^2+1^2} d x
\end{aligned}
$
$\begin{aligned}
= & \frac{1}{5}\left(\log \left|5 x^2+1\right|\right)_0^1+3 \cdot \frac{1}{1} \frac{\left(\tan ^{-1} \frac{\sqrt{5} x}{1}\right)_0^1}{\sqrt{5} \rightarrow \text { Coeff. of } x} \\
= & \frac{1}{5}\left(\log \left|5(1)^2+1\right|-\log \mid 5(0)^2+1\right)+\frac{3}{\sqrt{5}}\left(\tan ^{-1} \sqrt{5}-\tan ^{-1} \sqrt{0}\right) \\
= & \frac{1}{5}(\log 6-\log 1)+\frac{3}{\sqrt{5}}\left(\tan ^{-1} \sqrt{5}-0\right) \\
= & \frac{1}{5}(\log 6-0)+\frac{3}{\sqrt{5}}\left(\tan ^{-1} \sqrt{5}-0\right)
\end{aligned}$
$=\frac{1}{5} \log 6+\frac{3}{\sqrt{5}} \tan ^{-1} \sqrt{5} \text { Ans. }$
Ex 7.8 Question 15.
$\int_0^1 x e^{x^2} d x$
Answer.
$\int_0^1 x e^{x^2} d x$
First we evaluate $\int x e^{x^{x^2}} d x=\frac{1}{2} \int e^{x^{x^2}}(2 x) d x$ $\qquad$
Putting $x^2=t$
$
\begin{aligned}
& \Rightarrow 2 x=\frac{d t}{d x} \\
& \Rightarrow 2 x d x=d t \\
& \therefore \text { From eq. (i), } \frac{1}{2} \int e^t d t=\frac{1}{2} e^t=\frac{1}{2} e^{x^2} \\
& \therefore \int_0^1 x e^{x^2} d x=\frac{1}{2}\left(e^{x^2}\right)_0^1
\end{aligned}
$
$\begin{aligned}
& =\frac{1}{2}\left(e^1-e^0\right) \\
& =\frac{1}{2}(e-1) \text { Ans. }
\end{aligned}$
Ex 7.8 Question 16
$\int_1^2 \frac{5 x^2}{x^2+4 x+3} d x$
Answer.
$\int_1^2 \frac{5 x^2}{x^2+4 x+3} d x$
$\begin{aligned}
& =\int_1^2 \frac{5 x^2}{(x+1)(x+3)} d x \\
& =\int_1^2\left(5+\frac{-20 x-15}{(x+1)(x+3)}\right) d x \text { (On dividing) } \\
& =\int_1^2 5 d x+\int_1^2\left(\frac{-20 x-15}{(x+1)(x+3)}\right) d x \\
& =5(x)_1^2+\mathrm{I} \text { where } \mathrm{I}=\int_1^2\left(\frac{-20 x-15}{(x+1)(x+3)}\right) d x \\
& =5(2-1)+\mathrm{I}=5+\mathrm{I} \quad \text {.........(i) }
\end{aligned}$
Now, $\mathrm{I}=\int_1^2\left(\frac{-20 x-15}{(x+1)(x+3)}\right) d x$
Let $\frac{-20 x-15}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3}$
$
\begin{aligned}
& \Rightarrow-20 x-15=\mathrm{A}(x+3)+\mathrm{B}(x+1) \\
& \Rightarrow-20 x-15=\mathrm{A} x+3 \mathrm{~A}+\mathrm{B} x+\mathrm{B}
\end{aligned}
$
Comparing coefficients of $x \quad \mathrm{~A}+\mathrm{B}=-20$
Comparing constants $\quad 3 \mathrm{~A}+\mathrm{B}=-15$ $\qquad$ (v)
On solving eq. (iv) and (v), we get $A=\frac{5}{2}, B=\frac{-45}{2}$
Putting these values in eq. (iii),
$
\frac{-20 x-15}{(x+1)(x+3)}=\frac{5 / 2}{x+1}+\frac{-45 / 2}{x+3}
$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int_1^2\left(\frac{-20 x-15}{(x+1)(x+3)}\right) d x \\
& =\frac{5}{2} \int_1^2 \frac{1}{x+1} d x-\frac{45}{2} \int_1^2 \frac{1}{x+3} d x \\
& =\frac{5}{2}(\log |x+1|)_1^2-\frac{45}{2}\left(\left.\log |x+3|\right|_1 ^2\right. \\
& =\frac{5}{2}(\log |3|-\log |2|)-\frac{45}{2}(\log |5|-\log |4|) \\
& =\frac{5}{2} \log \frac{3}{2}-\frac{45}{2} \log \frac{5}{4} \\
& =\frac{5}{2}\left(\log \frac{3}{2}-9 \log \frac{5}{4}\right)
\end{aligned}
$
Putting this value of $\mathrm{I}$ in eq. (i),
$
\begin{aligned}
& \int_1^2 \frac{5 x^2}{x^2+4 x+3} d x \\
& =5+\frac{5}{2}\left(\log \frac{3}{2}-9 \log \frac{5}{4}\right) \\
& =5-\frac{5}{2}\left(9 \log \frac{5}{4}-\log \frac{3}{2}\right) \text { Ans. }
\end{aligned}
$
Ex 7.8 Question 17.
$\int_0^{\frac{\pi}{4}}\left(2 \sec ^2 x+x^3+2\right) d x$
Answer.
$\int_0^{\frac{\pi}{4}}\left(2 \sec ^2 x+x^3+2\right) d x$
$\begin{aligned}
& =2 \int_0^{\frac{\pi}{4}} \sec ^2 x d x+\int_0^{\frac{\pi}{4}} x^3 d x+2 \int_0^{\frac{\pi}{4}} 1 d x \\
& =2(\tan x)_0^{\frac{\pi}{4}}+\left(\frac{x^4}{4}\right)_0^{\frac{\pi}{4}}+2(x)_0^{\frac{\pi}{4}} \\
& =2\left(\tan \frac{\pi}{4}-\tan 0^{\circ}\right)+\frac{\left(\frac{\pi}{4}\right)^4}{4}-0+2\left(\frac{\pi}{4}-0\right) \\
& =2(1-0)+\frac{\left(\frac{\pi^4}{256}\right)}{4}+\frac{2 \pi}{4} \\
& =2+\frac{\pi^4}{1024}+\frac{\pi}{2} \\
& =\frac{\pi^4}{1024}+\frac{\pi}{2}+2 \text { Ans. }
\end{aligned}$
Ex 7.8 Question 18.
$\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x$
Answer.
$\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x$
$
\begin{aligned}
& =\int_0^\pi\left(\frac{1-\cos x}{2}-\frac{1+\cos x}{2}\right) d x \\
& =\int_0^\pi\left(\frac{1-\cos x-1-\cos x}{2}\right) d x \\
& =\int_0^\pi\left(\frac{-2 \cos x}{2}\right) d x
\end{aligned}
$
$\begin{aligned}
& =-\int_0^\pi \cos x d x \\
& =-(\sin x)_0^\pi \\
& =-\left(\sin \pi-\sin 0^{\circ}\right) \\
& =-(0-0)=0 \text { Ans. }
\end{aligned}$
Ex 7.8 Question 19.
$\int_0^2 \frac{6 x+3}{x^2+4} d x$
Answer.
$\int_0^2 \frac{6 x+3}{x^2+4} d x$
$
\begin{aligned}
& =\int_0^2\left(\frac{6 x}{x^2+4}+\frac{3}{x^2+4}\right) d x \\
& =\int_0^2 \frac{6 x}{x^2+4} d x+\int_0^2 \frac{3}{x^2+4} d x \\
& =3 \int_0^2 \frac{2 x}{x^2+4} d x+3 \frac{1}{2}\left(\tan ^{-1} \frac{x}{2}\right)_0^2 \\
& =3\left(\log \left|x^2+4\right|_0^2+\frac{3}{2}\left(\tan ^{-1} 1-\tan ^{-1} 0\right)\right. \\
& =3(\log 8-\log 4)+\frac{3}{2}\left(\frac{\pi}{4}-0\right) \\
& =3 \log \frac{8}{4}+\frac{3 \pi}{8} \\
& =3 \log 2+\frac{3 \pi}{8} \text { Ans. }
\end{aligned}
$
Ex 7.8 Question 20.
$\int_0^1\left(x e^2+\sin \frac{\pi x}{4}\right) d x$
Answer.
$\int_0^1\left(x e^x+\sin \frac{\pi x}{4}\right) d x$
$
=\int_0^1 x e^x d x+\int_0^1 \sin \frac{\pi x}{4} d x
$
[Applying Product Rule on first definite integral]
$
\begin{aligned}
& =\left(x e^x\right)_0^1-\int_0^1 1 e^x d x-\frac{\left(\cos \frac{\pi x}{4}\right)_0^1}{\frac{\pi}{4} \rightarrow \text { Coeff. of } x \text { in } \frac{\pi}{4}} \\
& =e^1-0-\int_0^1 e^x d x-\frac{4}{\pi}\left[\cos \frac{\pi}{4}-\cos 0^{\circ}\right] \\
& =e-\left(e^x\right)_0^1-\frac{4}{\pi}\left(\frac{1}{\sqrt{2}}-1\right) \\
& =e-\left(e-e^0\right)-\frac{4}{\pi \sqrt{2}}+\frac{4}{\pi} \\
& =e-e+1-\frac{2.2}{\pi \sqrt{2}}+\frac{4}{\pi} \\
& =1+\frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi} \text { Ans. }
\end{aligned}
$
Ex 7.8 Question 21.
$\int_1^{\sqrt{3}} \frac{d x}{1+x^2}$ equals:
(A) $\frac{\pi}{3}$
(B) $\frac{2 \pi}{3}$
(C) $\frac{\pi}{6}$
(D) $\frac{\pi}{12}$
Answer.
$\int_1^{\sqrt{9}} \frac{d x}{1+x^2}$
$
\begin{aligned}
& =\left(\tan ^{-1} x\right)_1^{\sqrt{6}} \\
& =\tan ^{-1} \sqrt{3}-\tan ^{-1} 1 \\
& =\frac{\pi}{3}-\frac{\pi}{4} \\
& =\frac{\pi}{12}
\end{aligned}
$
Therefore, option (D) is correct.
Ex 7.8 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.
$\int_0^{\frac{2}{3}} \frac{d x}{4+9 x^2}$
$
\begin{aligned}
& =\int_0^{\frac{2}{3}} \frac{d x}{(3 x)^2+2^2} \\
& =\left[\frac{1}{2} \frac{\tan ^{-1} \frac{3 x}{2}}{3 \rightarrow \text { Coeff. of } x \text { in } 3 x}\right]
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{6}\left[\tan ^{-1} \frac{3 x}{2}\right]_0^{\frac{2}{3}} \\
& =\frac{1}{6}\left[\tan ^{-1}\left(\frac{3}{2} \times \frac{2}{3}\right)-\tan ^{-1} 0\right] \\
& =\frac{1}{6}\left(\tan ^{-1} 1-\tan ^{-1} 0\right) \\
& =\frac{1}{6}\left(\frac{\pi}{4}-0\right)=\frac{\pi}{24}
\end{aligned}
$
$\text { Therefore, option (C) is correct. }$
