Anand Classes provides comprehensive NCERT Solutions and study material for Integrals Exercise 7.8 of Chapter 7 for Class 12 Mathematics, helping students understand advanced integration concepts through clear, step-by-step explanations. This Set-2 study material is aligned with the latest NCERT syllabus and is designed to support effective revision, concept mastery, and confident exam preparation. Click the print button to download study material and notes.
NCERT Question 12: Evaluate the integral
$$\int_{0}^{\dfrac{\pi}{2}}\cos^{2}x\;dx$$
Solution
Use the double-angle identity:
$$\cos^{2}x=\frac{1+\cos 2x}{2}$$
So the integral becomes
$$\int_{0}^{\dfrac{\pi}{2}}\cos^{2}x\;dx
=\frac{1}{2}\int_{0}^{\dfrac{\pi}{2}}\bigl(1+\cos 2x\bigr)\;dx$$
Find the antiderivative:
$$\int\bigl(1+\cos 2x\bigr)\;dx
=x+\frac{1}{2}\sin 2x + C$$
Apply the limits $0$ to $\frac{\pi}{2}$:
$$\frac{1}{2}\left[\;x+\frac{1}{2}\sin 2x\;\right]_{0}^{\dfrac{\pi}{2}}
=\frac{1}{2}\left(\frac{\pi}{2}+0-0-0\right)=\displaystyle \frac{\pi}{4}$$
Therefore the value is
$$\boxed{\displaystyle \frac{\pi}{4}}$$
Sharpen your calculus skills with clear, exam-ready notes from Anand Classes — download structured solution sheets and practice sets ideal for NCERT, CBSE and JEE revision.
NCERT Question 13: Evaluate the integral
$$\int_{2}^{3}\frac{x}{x^{2}+1}\;dx$$
Solution
Let
$$I=\int_{2}^{3}\frac{x}{x^{2}+1}\;dx$$
Rewrite the integrand:
$$\frac{x}{x^{2}+1}
=\frac{1}{2}\cdot\frac{2x}{x^{2}+1}$$
So,
$$I=\frac{1}{2}\int_{2}^{3}\frac{2x}{x^{2}+1}\;dx$$
Now integrate:
$$\int\frac{2x}{x^{2}+1}\;dx=\log(x^{2}+1)+C$$
Thus,
$$I=\frac{1}{2}\Bigl[\;\log(x^{2}+1)\;\Bigr]_{2}^{3}$$
Apply limits:
$$I=\frac{1}{2}\left[\log(3^{2}+1)-\log(2^{2}+1)\right]$$
$$I=\frac{1}{2}\left[\log(10)-\log(5)\right]$$
$$I=\frac{1}{2}\log\left(\frac{10}{5}\right)$$
$$I=\frac{1}{2}\log 2$$
So the value of the integral is
$$\boxed{\displaystyle \frac{1}{2}\log 2}$$
Enhance your preparation with detailed NCERT and JEE-focused solutions from Anand Classes — a perfect resource for CBSE boards, competitive exams, and concept strengthening.
NCERT Question 14: Evaluate the integral
$$\displaystyle \int_{0}^{1}\frac{2x+3}{5x^{2}+1}\;dx$$
Solution
$$\displaystyle \int_{0}^{1}\frac{2x+3}{5x^{2}+1}\;dx$$
Split the integrand:
$$\frac{2x+3}{5x^{2}+1}=\frac{2x}{5x^{2}+1}+\frac{3}{5x^{2}+1}.$$
For the first part use $u=5x^{2}+1,$ so $du=10x\;dx$ and $2x\;dx=\dfrac{1}{5}du$. Thus
$$\int\frac{2x}{5x^{2}+1}\;dx=\frac{1}{5}\int\frac{du}{u}=\frac{1}{5}\ln(5x^{2}+1).$$
For the second part use the standard formula $\displaystyle\int\frac{dx}{a x^{2}+1}=\frac{1}{\sqrt{a}}\tan^{-1}(x\sqrt{a})$. With $a=5$,
$$\int\frac{3}{5x^{2}+1}\;dx=\frac{3}{\sqrt{5}}\tan^{-1}(x\sqrt{5}).$$
Hence an antiderivative is
$$F(x)=\frac{1}{5}\ln(5x^{2}+1)+\frac{3}{\sqrt{5}}\tan^{-1}(x\sqrt{5}).$$
Evaluate from $0$ to $1$:
$$
\int_{0}^{1}\frac{2x+3}{5x^{2}+1}\;dx =F(1)-F(0) $$
$$\int_{0}^{1}\frac{2x+3}{5x^{2}+1}\;dx=\frac{1}{5}\ln(6)+\frac{3}{\sqrt{5}}\tan^{-1}(\sqrt{5})- \bigl(0+0\bigr)$$
$$\int_{0}^{1}\frac{2x+3}{5x^{2}+1}\;dx=\frac{1}{5}\ln 6+\frac{3}{\sqrt{5}}\tan^{-1}\bigl(\sqrt{5}\bigr).
$$
Final Answer
$$\boxed{\displaystyle \int_{0}^{1}\frac{2x+3}{5x^{2}+1}\;dx=\frac{1}{5}\ln 6+\frac{3}{\sqrt{5}}\tan^{-1}\bigl(\sqrt{5}\bigr)}$$
Boost your revision with downloadable, neatly formatted solution sets from Anand Classes — perfect for quick NCERT review and strong CBSE/JEE preparation.
NCERT Question 15: Evaluate the integral
$$\int_{0}^{1} x e^{x^{2}}\;dx$$
Solution
Let
$$I=\int_{0}^{1} x e^{x^{2}}\;dx$$
Substitute $t=x^{2}$ so that $dt=2x\;dx$ and $x\;dx=\tfrac{1}{2}dt$. When $x=0\Rightarrow t=0$ and $x=1\Rightarrow t=1$. Thus
$$I=\frac{1}{2}\int_{0}^{1} e^{t}\;dt.$$
Integrate:
$$I=\frac{1}{2}\bigl[e^{t}\bigr]_{0}^{1}=\frac{1}{2}\bigl(e-1\bigr).$$
Final Answer
$$\boxed{\displaystyle \int_{0}^{1} x e^{x^{2}}\;dx=\frac{1}{2}\bigl(e-1\bigr)}$$
Download concise carefully worked notes by Anand Classes — perfect for JEE and CBSE preparation and ideal as quick revision material for integration and partial fractions.
NCERT Question 16: Evaluate the integral
$$I=\int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}dx$$
Solution
$$I=\int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}dx$$
First perform polynomial division, rewrite the integrand:
$$\frac{5x^{2}}{x^{2}+4x+3}=5-\frac{20x+15}{x^{2}+4x+3}$$
factor the denominator:
$$x^{2}+4x+3=(x+1)(x+3)$$
do partial fractions for the remainder:
$$\frac{20x+15}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3}$$
Equating coefficients gives
$$A+B=20,\qquad 3A+B=15$$
so
$$2A=-5\implies A=-\frac{5}{2},\qquad B=20-A=\frac{45}{2}$$
Thus the integrand becomes
$$\frac{5x^{2}}{x^{2}+4x+3}=5+\frac{5/2}{x+1}-\frac{45/2}{x+3}$$
Integrate termwise:
$$I=\int_{1}^{2}\left(5+\frac{5/2}{x+1}-\frac{45/2}{x+3}\right)dx$$
$$I=\left[5x+\frac{5}{2}\ln|x+1|-\frac{45}{2}\ln|x+3|\right]_{1}^{2}$$
Evaluate at the limits:
At $x=2$:
$$5x+\frac{5}{2}\ln(x+1)-\frac{45}{2}\ln(x+3)=10+\frac{5}{2}\ln 3-\frac{45}{2}\ln 5$$
At $x=1$:
$$5x+\frac{5}{2}\ln(x+1)-\frac{45}{2}\ln(x+3)=5+\frac{5}{2}\ln 2-\frac{45}{2}\ln 4$$
Subtract:
$$I=\left[5x+\frac{5}{2}\ln|x+1|-\frac{45}{2}\ln|x+3|\right]_{1}^{2}=5+\frac{5}{2}\ln\frac{3}{2}-\frac{45}{2}\ln\frac{5}{4}$$
Factor $\tfrac{5}{2}$ if desired:
$$I=5+\frac{5}{2}\left(\ln\frac{3}{2}-9\ln\frac{5}{4}\right)$$
Final answer
$$\boxed{I=5+\frac{5}{2}\ln\frac{3}{2}-\frac{45}{2}\ln\frac{5}{4},}$$
NCERT Question.17 : Evaluate the integral
$$I=\int_{0}^{\dfrac{\pi}{4}}\left(2\sec^{2}x+x^{3}+2\right)dx$$
Solution
$$I=\int_{0}^{\dfrac{\pi}{4}}\left(2\sec^{2}x+x^{3}+2\right)dx$$
Find an antiderivative termwise. An antiderivative of the integrand is
$$F(x)=2\tan x+\frac{x^{4}}{4}+2x$$
so by the Fundamental Theorem of Calculus
$$I=F\left(\frac{\pi}{4}\right)-F(0)$$
Evaluate at the limits:
$$F\left(\frac{\pi}{4}\right)=2\tan\frac{\pi}{4}+\frac{\left(\frac{\pi}{4}\right)^{4}}{4}+2\cdot\frac{\pi}{4}
=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}$$
$$F(0)=2\tan 0+0+0=0$$
Therefore
$$I=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}$$
Final answer
$$\boxed{I=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024},}$$
Download notes by Anand Classes — concise solutions perfect for JEE and CBSE preparation and ideal for quick revision and practice.
NCERT Question 18: Evaluate the integral
$$I=\int_{0}^{\pi}\left(\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}\right)dx$$
Solution
$$I=\int_{0}^{\pi}\left(\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}\right)dx$$
Use the identity
$$\sin^{2}\theta-\cos^{2}\theta=-\cos2\theta$$
with $\theta=\dfrac{x}{2}$, giving
$$\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}=-\cos x$$
Thus,
$$I=\int_{0}^{\pi}\left(\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}\right)dx
=-\int_{0}^{\pi}\cos x\;dx$$
An antiderivative of $\cos x$ is $\sin x$, so
$$I=-[\sin x]_{0}^{\pi}$$
Evaluate the limits:
$$I=-(\sin\pi-\sin0)=-(0-0)=0$$
Final answer
$$\boxed{I=0}$$
High-quality step-by-step solutions by Anand Classes — ideal for CBSE and JEE preparation, and perfect for strengthening your calculus fundamentals.
NCERT Question 19: Evaluate the integral
$$I=\int_{0}^{2}\frac{6x+3}{x^{2}+4}dx$$
Solution
Let $$I=\int_{0}^{2}\frac{6x+3}{x^{2}+4}dx$$
Write the integrand as a constant factor times a sum:
$$\frac{6x+3}{x^{2}+4}=3\cdot\frac{2x+1}{x^{2}+4} =3\left(\frac{2x}{x^{2}+4}+\frac{1}{x^{2}+4}\right)$$
Integrate termwise. An antiderivative is
$$F(x)=3\left(\ln\big(x^{2}+4\big)+\frac{1}{2}\tan^{-1}\frac{x}{2}\right)$$
so by the Fundamental Theorem of Calculus
$$I=F(2)-F(0)$$
Evaluate the limits:
$$F(2)=3\left(\ln(8)+\frac{1}{2}\tan^{-1}1\right)=3\ln 8+\frac{3}{2}\cdot\frac{\pi}{4}=3\ln 8+\frac{3\pi}{8}$$
$$F(0)=3\left(\ln 4+\frac{1}{2}\tan^{-1}0\right)=3\ln 4+0=3\ln 4$$
Therefore
$$I=3\ln 8-3\ln 4+\frac{3\pi}{8}=3\ln\frac{8}{4}+\frac{3\pi}{8}=3\ln 2+\frac{3\pi}{8}$$
Final answer
$$\boxed{I=3\ln 2+\frac{3\pi}{8}}$$
Concise, exam-focused notes by Anand Classes — clear worked solutions ideal for JEE and CBSE revision and quick practice.
NCERT Question 20: Evaluate the integral
$$I=\int_{0}^{1}\left(xe^{x}+\sin\left(\frac{\pi x}{4}\right)\right)dx$$
Solution
$$I=\int_{0}^{1}\left(xe^{x}+\sin\left(\frac{\pi x}{4}\right)\right)dx$$
Split the integral:
$$I=\int_{0}^{1}xe^{x}dx+\int_{0}^{1}\sin\left(\frac{\pi x}{4}\right)dx$$
First integral:
$$\int xe^{x}dx=xe^{x}-e^{x}=e^{x}(x-1)$$
Evaluate from (0) to (1):
$$\left[e^{x}(x-1)\right]_{0}^{1}=e(1-1)-1(0-1)=1$$
Second integral:
Let
$$\int\sin\left(\frac{\pi x}{4}\right)dx$$
The antiderivative is
$$-\frac{4}{\pi}\cos\left(\frac{\pi x}{4}\right)$$
Evaluate from (0) to (1):
$$\left[-\frac{4}{\pi}\cos\left(\frac{\pi x}{4}\right)\right]_{0}^{1}
=-\frac{4}{\pi}\cos\left(\frac{\pi}{4}\right)+\frac{4}{\pi}\cos 0$$
$$=-\frac{4}{\pi}\left(\frac{\sqrt{2}}{2}\right)+\frac{4}{\pi}
=\frac{4}{\pi}\left(1-\frac{\sqrt{2}}{2}\right)$$
Add both parts:
$$I=1+\frac{4}{\pi}\left(1-\frac{\sqrt{2}}{2}\right)$$
Final answer
$$\boxed{I=1+\frac{4}{\pi}\left(1-\frac{\sqrt{2}}{2}\right)}$$
Get high-quality integral solutions and study material from Anand Classes — perfect for JEE and CBSE students who want clear, step-by-step explanations for fast revision.
NCERT Question 21: Evaluate the integral
$$\int_{1}^{\sqrt{3}}\frac{dx}{1+x^{2}}$$
Solution
$$\int_{1}^{\sqrt{3}}\frac{dx}{1+x^{2}}$$
Use the standard formula
$$\int \frac{dx}{1+x^{2}}=\tan^{-1}x$$
Evaluate the definite integral:
$$I=\left[\tan^{-1}x\right]_{1}^{\sqrt{3}}$$
$$I=\tan^{-1}(\sqrt{3})-\tan^{-1}(1)$$
Since
$$\tan^{-1}(1)=\frac{\pi}{4}$$
and
$$\tan^{-1}(\sqrt{3})=\frac{\pi}{3}$$
(for the given options)
Therefore,
$$I=\frac{\pi}{3}-\frac{\pi}{4}
=\frac{4\pi-3\pi}{12}
=\frac{\pi}{12}$$
Final answer
$$\boxed{\frac{\pi}{12}}$$
Trusted step-by-step solutions by Anand Classes — ideal for JEE and CBSE practice, helping students build strong conceptual clarity in calculus.
NCERT Question 22: Evaluate the integral
$$I=\int_{0}^{\dfrac{2}{3}}\frac{dx}{4+9x^{2}}$$
Solution
$$I=\int_{0}^{\dfrac{2}{3}}\frac{dx}{4+9x^{2}}$$
Write the integrand in standard form:
$$\frac{1}{4+9x^{2}}=\frac{1}{4\left(1+\left(\frac{3x}{2}\right)^{2}\right)}$$
Thus,
$$I=\int_{0}^{\frac{2}{3}}\frac{1}{4\left(1+\left(\frac{3x}{2}\right)^{2}\right)}dx
=\frac{1}{4}\int_{0}^{\frac{2}{3}}\frac{dx}{1+\left(\frac{3x}{2}\right)^{2}}$$
Let
$$u=\frac{3x}{2}\Rightarrow du=\frac{3}{2}dx\Rightarrow dx=\frac{2}{3}du$$
When $(x=0\Rightarrow u=0)$
When $(x=\frac{2}{3}\Rightarrow u=1)$
Now substitute:
$$I=\frac{1}{4}\int_{0}^{1}\frac{1}{1+u^{2}}\left(\frac{2}{3}\right)du$$
$$I=\frac{1}{6}\int_{0}^{1}\frac{du}{1+u^{2}}$$
The antiderivative is:
$$\int\frac{du}{1+u^{2}}=\tan^{-1}u$$
So,
$$I=\frac{1}{6}\left[\tan^{-1}u\right]_{0}^{1}$$
$$I=\frac{1}{6}\left(\tan^{-1}1-\tan^{-1}0\right)$$
$$I=\frac{1}{6}\left(\frac{\pi}{4}-0\right)$$
$$I=\frac{\pi}{24}$$
Final answer
$$\boxed{\frac{\pi}{24}}$$
Download high-quality notes and solved integrals from Anand Classes — perfect for JEE and CBSE students who want clear, accurate, and exam-focused solutions.
