CBSE Class 12 Maths Chapter 7 Integrals Important Questions
Question 1. Find \(\int e^x\left(\log \sqrt{x}+\frac{1}{2 x}\right) d x\) Or,
Find \(\int \mathrm{e}^{2 \log x} \mathrm{dx}\)
Solution:
Let I = \(\int e^x\left(\log \sqrt{x}+\frac{1}{2 x}\right) d x\)
Let f(x) = \(\log \sqrt{x}\)
⇒ \(f^{\prime}(x)=\frac{1}{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}=\frac{1}{2 x}\)
⇒ \(\mathrm{I}=\mathrm{e}^{\mathrm{x}} \log \sqrt{\mathrm{x}}+\mathrm{c}\) (because \(\int \mathrm{e}^{\mathrm{x}}\left(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right) \mathrm{dx}=\mathrm{e}^{\mathrm{x}} \cdot \mathrm{f}(\mathrm{x})+\mathrm{c}\))
Or,
∴ I = \(\int e^{2 \log x} d x\)
I = \(\int e^{\log x^2} d x=\int x^2 d x\) (because \(e^{\log x}=x\))
= \(\frac{x^3}{3}+C\)
Read and Learn More CBSE Class 12 Maths Important Question and Answers
Question 2. Evalute: \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos ^2 x d x\)
Solution:
Let I = \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos ^2 x d x\)
Here f(x) = x.cos²x
f(-x) = (-x) cos²(-x)
f(-x) = -x cos²x
f(-x) = -f(x)
i.e. f(x) is odd function
∴ I = 0 (\(\int_{-a}^a\) f(x)dx = 0, if f(x) is odd function)
Class 12 Maths Integrals Solved Examples
Question 3. Find: \(\int \frac{d x}{x^2-6 x+13}\)
Solution:
I = \(\int \frac{d x}{x^2-6 x+13}=\int \frac{1}{(x-3)^2+2^2} d x\)
= \(\frac{1}{2} \tan ^{-1}\left(\frac{x-3}{2}\right)+C\)
(because \(\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\))
Question 4. Find: \(\int \frac{\tan ^3 x}{\cos ^3 x} d x\)
Solution:
Let \(I=\int \frac{\tan ^3 x}{\cos ^3 x} d x=\int \frac{\sin ^3 x}{\cos ^6 x} d x=\int \frac{\sin ^2 x \cdot \sin x}{\cos ^6 x} d x=\int \frac{\left(1-\cos ^2 x\right) \cdot \sin x}{\cos ^6 x} d x\)
Put \(\cos x=t \Rightarrow-\sin x d x=d t \Rightarrow \sin x d x=-d t\)
Now I = \(-\int \frac{\left(1-t^2\right)}{t^6} d t=-\int\left(\frac{1}{t^6}-\frac{1}{t^4}\right) d t=-\left[\frac{-1}{5 t^5}+\frac{1}{3 t^3}\right]+C=\frac{1}{5 \cos ^5 x}-\frac{1}{3 \cos ^3 x}+C\)
Question 5. Find: \(\int \frac{x-5}{(x-3)^3} e^x d x\)
Solution:
⇒ \(\int \frac{x-5}{(x-3)^3} e^x d x =\int e^x\left[\frac{(x-3)-2}{(x-3)^3}\right] d x=\int e^x\left[\frac{(x-3)}{(x-3)^3}-\frac{2}{(x-3)^3}\right] d x\)
= \(\int e^x\left[\frac{1}{(x-3)^2}+\left\{\frac{-2}{(x-3)^3}\right\}\right] d x\) [because \(\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+C)\)
= \(\frac{e^x}{(x-3)^2}+C\)
Question 6. Find: \(\int \frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x} d x\) Or,
Find: \(\int \frac{x-3}{(x-1)^3} e^x d x\)
Solution:
⇒ \(\int \frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cdot \cos ^2 x} d x=\int\left(\frac{\sin x}{\cos ^2 x}+\frac{\cos x}{\sin ^2 x}\right) d x\)
= \(\int(\tan x \sec x+\mathrm{cosec} x \cdot \cot x) d x=\sec x-\mathrm{cosec} x+C\)
Or
⇒ \(\int \frac{x-3}{(x-1)^3} e^x d x=\int e^x\left(\frac{x-1-2}{(x-1)^3}\right) d x\)
= \(\int \mathrm{e}^x\left(\frac{1}{(\mathrm{x}-1)^2}+\left(\frac{-2}{(\mathrm{x}-1)^3}\right)\right) \mathrm{dx}\)
(because \(\int \mathrm{e}^{\mathrm{x}}\left(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right) \mathrm{dx}=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})+C\)
= \(\frac{e^x}{(x-1)^2}+C\)
Definite Integrals Problems For Class 12 Maths CBSE
Question 7. Evalute \(\int \frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x} d x\)
Solution:
⇒ \(\int \frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x} d x=\int \frac{1-2 \sin ^2 x+2 \sin ^2 x}{\cos ^2 x} d x\)
= \(\int \frac{1}{\cos ^2 x} d x=\int \sec ^2 x d x=\tan x+C\)
Question 8. Evaluate \(\int_0^{2 \pi} \frac{\mathrm{dx}}{1+\mathrm{e}^{\sin x}}\)
Solution:
Let I = \(\int_0^{2 \pi} \frac{1}{1+e^{\sin x}} d x\)
⇒ I = \(\int_0^{2 \pi} \frac{1}{1+e^{\sin (2 \pi-x)}} d x\) (\(because \int_0^a f(x) d x=\int_0^a f(a-x) d x\))
⇒ I = \(\int_0^{2 \pi} \frac{1}{1+e^{-\sin x}} d x \quad \Rightarrow I=\int_0^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x\)
From (1) and (2), we get :
⇒ 2I = \(\int_0^{2 \pi} \frac{1}{1+e^{\sin x}} d x+\int_0^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x \Rightarrow 2 I=\int_0^{2 \pi} 1 \cdot d x\)
⇒ 2I = \([x]_0^{2 \pi} \Rightarrow 2 I=2 \pi \Rightarrow I=\pi\)
Question 9. Evaluate: \(\int_{-1}^2\left|x^3-x\right| d x\)
Solution:
We know that x³ – x ≥ 0 on [0, 1] and x³ – x ≤ 0 on [0, 1] and that x³ – x ≥ 0 on [1, 2], so by property of definite integral we get
⇒ \(\int_{-1}^2\left|x^3-x\right| d x=\int_{-1}^0\left(x^3-x\right) d x+\int_0^1-\left(x^3-x\right) d x+\int_1^2\left(x^3-x\right) d x\)
= \(\int_{-1}^0\left(x^3-x\right) d x+\int_0^1\left(x-x^3\right) d x+\int_1^2\left(x^3-x\right) d x\)
= \(\left[\frac{x^4}{4}-\frac{x^2}{4}\right]_{-1}^0+\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1+\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_1^2\)
= \(-\left(\frac{1}{4}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+(4-2)-\left(\frac{1}{4}-\frac{1}{2}\right)\)
= \(-\frac{1}{4}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+2-\frac{1}{4}+\frac{1}{2}=\frac{3}{2}-\frac{3}{4}+2=\frac{11}{4}\)
Question 10. Evaluate: \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\sin |x|+\cos |x|) d x\)
Solution:
Let \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\sin |x|+\cos |x|) d x\)
∵ f(x) == sin |x| + cos |x|
f(-x) = sin |-x| + cos |-x|
f(-x) = sin |x| + cos |x| = f(x)
∴ f(x) is even function
Therfore by prop \(\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x\)
I = \(2 \int_0^{\pi / 2}\{\sin |x|+\cos |x|\} d x\)
I = \(2 \int_0^{\pi / 2}\{\sin x+\cos x\} d x\)
I = \(2[-\cos x+\sin x]_0^{\pi / 2}\)
I = \(2[0+1-(-1)-0]=2 \times 2=4\)
Question 11. Find: \(\int \frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)} d x\) or,
Evaluate : \(\int_{-2}^1 \sqrt{5-4 x-x^2} d x\)
Solution:
I = \(\int \frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)} d x\)
Let \(x^2=y\)
Let \(\frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)}=\frac{y}{(y+1)(3 y+4)}=\frac{A}{(y+1)}+\frac{B}{(3 y+4)}\)
⇒ \(\mathrm{y}=\mathrm{A}(3 \mathrm{y}+4)+\mathrm{B}(\mathrm{y}+1)\)
from equation (1)
Put y=-1 ⇒ A=-1
Put y=-4/3 ⇒ B=4
∴ \(\frac{y}{(y+1)(3 y+4)}=\frac{-1}{y+1}+\frac{4}{3 y+4}\)
i.e. \(\frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)}=\frac{-1}{x^2+1}+\frac{4}{3 x^2+4}\)
⇒ I = \(\int\left(\frac{-1}{x^2+1}+\frac{4}{3 x^2+4}\right) d x=-\int \frac{1}{x^2+1} d x+\frac{4}{3} \int \frac{1}{x^2+4 / 3} d x\)
= \(-\tan ^{-1} x+\frac{4}{3} \cdot \frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{x \sqrt{3}}{2}\right)+C\)
=\(\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x \sqrt{3}}{2}\right)-\tan ^{-1} x+C\)
Or,
Let I = \(\int_{-2}^1 \sqrt{5-4 x-x^2} d x\)
= \(\int_{-2}^1 \sqrt{3^2-(x+2)^2} d x\) (because \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c\)
= \(\left[\frac{x+2}{2} \sqrt{3^2-(x+2)^2}+\frac{9}{2} \sin ^{-1}\left(\frac{x+2}{3}\right)\right]_{-2}^1\)
= \(\frac{9}{2} \sin ^{-1} 1=\frac{9 \pi}{4}\)
Indefinite Integrals Questions Class 12 Maths
Question 12 Find \( \int\frac{x^2}{(x-1)(x+1)^2} d x\)
Solution:
let I = \( \int\frac{x^2}{(x-1)(x+1)^2} d x\)
Let \(\frac{x^2}{(x-1)(x+1)^2}=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2}\)
⇒ x³ = A(x + 1)² + B(x – 1 )(x + 1) + C(x – 1)
Put x = 1, ⇒ 1 = 4A ⇒ A = 1/4
Put x = -1, ⇒ 1 = C(-2) ⇒ C = -1/2
Comparing the coefficients of x² on both sides:
1 = A + B ⇒ B = 1 – A = 1-1/4 = 3/4
Hence, \(I=\frac{1}{4} \int \frac{1}{(x-1)} d x+\frac{3}{4} \int \frac{1}{(x+1)} d x-\frac{1}{2} \int \frac{1}{(x+1)^2} d x\)
I = \(\frac{1}{4} \log |x-1|+\frac{3}{4} \log |x+1|+\frac{1}{2(x+1)}+C\)
Question 13. Find
- \(\int \frac{x^2+2}{x^2+1} d x\)
- \(\int_{-1}^1 \frac{|x|}{x} d x\)
Or,
Evaluate \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x\)
Solution:
1. I = \(\int \frac{x^2+2}{x^2+1} d x=\int\left(\frac{x^2+1+1}{x^2+1}\right) d x=\int \frac{x^2+1}{x^2+1} d x+\int \frac{1}{x^2+1} d x\)
= \(\int 1 \cdot d x+\int \frac{1}{x^2+1} d x=x+\tan ^{-1} x+C\)
2. Let \(I=\int_{-1}^1 \frac{|x|}{x} d x\)
Now, \(\frac{|x|}{x}= \begin{cases}\frac{-x}{x}=-1, & x<0 \\ \frac{x}{x}=1 & , x \geq 0\end{cases}\)
⇒ \(\mathrm{I} =\int_{-1}^0(-1) \cdot \mathrm{dx}+\int_0^1 1 \cdot \mathrm{dx}=[-\mathrm{x}]_{-1}^0+[\mathrm{x}]_0^1=-[0+1]+[1-0]\)
=-1+1=0
Or,
Let I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x\)….(1)
Using property \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\);
I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right) d x\)
⇒ I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x\)….(2)
Adding (1) and (2), we get
2I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x+\int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x\)
2I =\(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x} \cdot \frac{4+3 \cos x}{4+3 \sin x}\right) d x\)
2I =\(\int_0^{\pi / 2} \log 1 d x \Rightarrow 2 I=0\) (because log 1=0)
⇒ I =0
Question 14. Find \(\int \frac{x^3+1}{x^3-x} d x\)
Solution:
I = \(\int \frac{x^3+1}{x^3-x} d x\)
I = \(\int \frac{x^3+1-x+x}{x^3-x} d x \text { or } I=\int\left(1+\frac{1+x}{x^3-x}\right) d x\)
= \(\int\left(1+\frac{1+x}{x\left(x^2-1\right)}\right) d x=\int\left(1+\frac{1}{x(x-1)}\right) d x\)
= \(\int\left(1+\frac{1}{x-1}-\frac{1}{x}\right) d x=x+\log |x-1|-\log |x|+C\)
= \(x+\log \left|\frac{x-1}{x}\right|+C\)
Graphical Questions On Integrals Class 12 CBSE
Question 15. Evaluate: \(\int \frac{2 \cos x}{(1-\sin x)\left(2-\cos ^2 x\right)} d x\)
Solution:
I = \(\int \frac{2 \cos x}{(1-\sin x)\left(2-\cos ^2 x\right)} d x\)=\(\int \frac{2 \cos x}{(1-\sin x)\left(1+1-\cos ^2 x\right)} d x\)=\(\int \frac{2 \cos x}{(1-\sin x)\left(1+\sin ^2 x\right)} d x\)
put \(\sin \mathrm{x}=\mathrm{t} \Rightarrow \cos \mathrm{dx}=\mathrm{dt}\)
= \(\int \frac{2 d t}{(1-t)\left(1+t^2\right)}\)
⇒ \(\mathrm{I}=-2 \mathrm{I_1}\)….(1)
Now; \(I_1=\int \frac{d t}{(t-1)\left(t^2+1\right)}\)
Let \(\frac{1}{(t-1)\left(t^2+1\right)}=\frac{A}{(t-1)}+\frac{B t+C}{\left(t^2+1\right)}\)
I =A(t²+1)+B t(t-1)+C(t-1)
Putting t = 1
⇒ 1 = A x 2
⇒ A = 1/2
Comparing the coefficient of ‘t²’ on both sides we get 0 = A + B
B = -1/2
Putting t = 0, 1 = A x 1 + C x (-1)
⇒ 1 = A – C
C = A – 1 = -1/2
⇒ \(I_1=\int \frac{d t}{(t-1)\left(t^2+1\right)}=\frac{1}{2} \int \frac{d t}{t-1}-\frac{1}{2} \int \frac{t+1}{t^2+1} d t\)
= \(\frac{1}{2} \log |t-1|-\frac{1}{4} \int \frac{2 t}{t^2+1} d t-\frac{1}{2} \int \frac{d t}{t^2+1}+C_1\)
⇒ \(I_1=\frac{1}{2} \log |t-1|-\frac{1}{4} \log \left|t^2+1\right|-\frac{1}{2} \tan ^{-1} t+C_1\)
I = \(-2 I_1=-2\left[\frac{1}{2} \log |t-1|-\frac{1}{4} \log \left|t^2+1\right|-\frac{1}{2} \tan ^{-1} t+C_1\right]\)
= \(-\log |t-1|+\frac{1}{2} \log \left|t^2+1\right|+\tan ^{-1} t-2 C_1\)
⇒ I =\(-\log |\sin x-1|+\frac{1}{2} \log \left|1+\sin ^2 x\right|+\tan ^{-1}(\sin x)+C, \text { where } C=-2 C_1\)
Question 16. Prove that \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a-x) dx and hence evaluate \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x} d x\)
Solution:
Let I = \(\int_0^a f(a-x) d x\)
Put \(\mathrm{a}-\mathrm{x}=\mathrm{t} \Rightarrow \mathrm{dx}=-\mathrm{dt}\)
When \(\mathrm{x}=0, \mathrm{t}=\mathrm{a} \& \mathrm{x}=\mathrm{a}, \mathrm{t}=0\)
⇒ I = \(-\int_a^0 f(t) d t\)
or I = \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{t}) \mathrm{dt}\)
(because \(\int_a^b f(x) dx=-\int_b^a f(x) d x\))
⇒ \(\mathrm{I}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\)
(because \(\int_a^b f(x) d x=\int_a^b f(t) d t\))
Now; I = \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x} d x\)….(1)
I = \(\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x\)
= \(\int_0^{\pi / 2} \frac{\frac{\pi}{2}-x}{\cos x+\sin x} d x\)….(2)
⇒ (1) + (2) gives:
2I = \(\int_0^{\pi / 2} \frac{\frac{\pi}{2}}{\cos x+\sin x} d x\)
∴ 2I = \(\frac{\pi}{2} \int_0^{\pi / 2} \frac{1}{\sin x+\cos x} d x\)
⇒ I = \(\frac{\pi}{4 \sqrt{2}} \int_0^{\pi / 2} \frac{1}{\sin \left(x+\frac{\pi}{4}\right)} d x\)
Or, I = \(\frac{\pi}{4 \sqrt{2}} \int_0^{\pi / 2} \mathrm{cosec}\left(x+\frac{\pi}{4}\right) d x\)
⇒ I = \(\frac{\pi}{4 \sqrt{2}}\left[\log \left(\mathrm{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)\right]_0^{\pi / 2}\)
or I = \(\frac{\pi}{4 \sqrt{2}}\left[\log \left(\mathrm{cosec}\left(\frac{\pi}{2}+\frac{\pi}{4}\right)-\cot \left(\frac{\pi}{2}+\frac{\pi}{4}\right)\right)-\log \left(\mathrm{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right)\right]\)
∴ I = \(\frac{\pi}{4 \sqrt{2}}\left[\log \left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)-\log \left(\mathrm{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right)\right]\)
or I = \(\frac{\pi}{4 \sqrt{2}}[\log (\sqrt{2}+1)-\log (\sqrt{2}-1)] \Rightarrow I=\frac{\pi}{4 \sqrt{2}} \log \frac{\sqrt{2}+1}{\sqrt{2}-1}\)
Question 17. Evaluate: \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x} d x\)
Solution:
Let \(I=\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x} d x\)=\(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9\left\{1-(\sin x-\cos x)^2\right\}} d x\)
Put sin x – cos x = t ⇒ (cos x + sin x)dx = dl
when x = 0, t = -1 and x = π/4, t = 0
∴ I = \(\int_{-1}^0 \frac{\mathrm{dt}}{16+9\left(1-\mathrm{t}^2\right)}=\int_{-1}^0 \frac{\mathrm{dt}}{16+9-9 \mathrm{t}^2}\)
= \(\int_{-1}^0 \frac{\mathrm{dt}}{25-9 \mathrm{t}^2}=\frac{1}{9} \int_{-1}^0 \frac{\mathrm{dt}}{\frac{25}{9}-\mathrm{t}^2}=\frac{1}{9} \int_{-1}^0 \frac{\mathrm{dt}}{\left(\frac{5}{3}\right)^2-(\mathrm{t})^2}\)
(because \(\int \frac{1}{\mathrm{a}^2-\mathrm{x}^2} \mathrm{dx}=\frac{1}{2 \mathrm{a}} \log \left|\frac{\mathrm{a}+\mathrm{x}}{\mathrm{a}-\mathrm{x}}\right|+\mathrm{c})\)
= \(\frac{1}{9} \cdot \frac{1}{2} \times \frac{3}{5}\left[\log \left(\frac{\frac{5}{3}+\mathrm{t}}{\frac{5}{3}-\mathrm{t}}\right)\right]_{-1}^0\)
= \(\frac{1}{30}\left[\log (1)-\log \left(\frac{1}{4}\right)\right]=\frac{1}{30} \log 4\)