Integral Calculus Questions and Answers Part-3

1.If \[I=\int\frac{dx}{\sqrt{\left(15+2x+x^{2}\right)^{3}}},=A\frac{x+1}{\sqrt{5+2x+x^{2}}}+C\]
then A is equal to
a) \[\frac{1}{2}\]
b) 2
c) \[\frac{1}{4}\]
d) \[\frac{4}{3}\]

Answer: c
Explanation:
q21
q21a

2. If \[I=\int\frac{\cos x}{\sin\left(x-a\right)}dx\]
then I is equal to
a) \[\cos a\log\mid \sin\left(x-a\right)\mid-x\sin a+C\]
b) \[\cos a\log\mid \sin\left(x-a\right)\mid+x\sin a+C\]
c) \[-\cos a\log\mid \sin\left(x-a\right)\mid-x\sin a+C\]
d) \[x\sin a-\cos a\log \mid\sin\left( x-a\right)\mid+C\]

Answer: a
Explanation:
q22
q22a

3.If \[I=\int\left[1+2\tan x\left(\tan x+\sec x\right)\right]^{1/2}dx\]
then I is equal to
a) \[-\log \mid\sec x+\tan x )+\log \mid\cos x\mid+C\]
b) \[\log \mid\frac{\tan\left(x/2+\pi/4\right)}{\cos x}\mid+C\]
c) \[\log \mid\tan\left(x/2+\pi/4\right)\cos x\mid+C\]
d) \[\log \mid\cot\left(\pi/4-x/2\right)\cos x\mid+C\]

Answer: b
Explanation:
q23

4. If \[I=\int\frac{dx}{x^{4}\sqrt{a^{2}+x^{2}}}\]
then I is equal to
a) \[\frac{1}{a^{4}}\left[\frac{1}{x}\sqrt{a^{2}+x^{2}}-\frac{1}{3x^{3}}\sqrt{a^{2}+x^{2}}\right]+C\]
b) \[\frac{1}{a^{4}}\left[\frac{1}{x}\sqrt{a^{2}+x^{2}}-\frac{1}{3x^{3}}\left(a^{2}+x^{2}\right)^{3/2}\right]+C\]
c) \[\frac{1}{a^{4}}\left[\frac{1}{x}\sqrt{a^{2}+x^{2}}-\frac{1}{2\sqrt{x}}\left(a^{2}+x^{2}\right)^{3/2}\right]+C\]
d) \[\frac{2}{a^{4}}\left[\frac{1}{x}\sqrt{a^{2}+x^{2}}-\frac{1}{x^{3}}\sqrt{a^{2}+x^{2}}\right]+C\]

Answer: b
Explanation:
q24
q24a

5. If \[I=\int\frac{dx}{\sin^{4}x+\cos^{4}x}\]
then I is equal to
a) \[\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan 2x}{\sqrt{2}}\right)+C\]
b) \[\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{1+\cos 2x}{\sqrt{2}}\right)+C\]
c) \[\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x+\cot x}{\sqrt{2}}\right)+C\]
d) \[\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{2}}\right)+C\]

Answer: a
Explanation:
q25
q25a

6. If \[I=\int\frac{\sin^{4}x}{\sin^{4}x+\cos^{4}x}dx\]
then I is equal to
a) \[\frac{1}{2}\left[x+\frac{1}{2\sqrt{2}}\log \mid \frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\mid\right]+C\]
b) \[\frac{1}{2}\left[x-\frac{1}{2\sqrt{2}}\log \mid \frac{1+\sin 2x}{1-\sin 2x}\mid\right]+C\]
c) \[\frac{1}{2}\left[x-\frac{1}{2\sqrt{2}}\log \mid \frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\mid\right]+C\]
d) \[\frac{1}{2}\left[x+\frac{1}{2\sqrt{2}}\log \mid \frac{\sqrt{2+\sin 2x}}{\sqrt{2-\sin 2x}}\mid\right]+C\]

Answer: c
Explanation:
q26
q26a

7. If \[I=\int\sqrt{\frac{5-x}{2+x}}dx\]
then I is equal to
a) \[\sqrt{x+2}\sqrt{5-x}+3\sin^{-1}\sqrt{\frac{x+2}{3}}+C\]
b) \[\sqrt{x+2}\sqrt{5-x}+7\sin^{-1}\sqrt{\frac{x+2}{7}}+C\]
c) \[\sqrt{x+2}\sqrt{5-x}+5\sin^{-1}\sqrt{\frac{x+2}{5}}+C\]
d) \[\sqrt{10+3x-x^{2}}+3\sin^{-1}\sqrt{\frac{x+2}{5}}+C\]

Answer: b
Explanation:
q27

8. If \[I=\int\left(x+1\right)\sqrt{\frac{x+2}{x-2}}dx\]
then I is equal to
a) \[\frac{1}{2}\left(x+6\right)\sqrt{x^{2}-4}+2\log \mid x+\sqrt{x^{2}-4}\mid +C\]
b) \[\frac{1}{2}\left(x+6\right)\sqrt{x^{2}-4}+4\log \mid x+\sqrt{x^{2}-4}\mid +C\]
c) \[\frac{1}{2}\left(x+6\right)\sqrt{x^{2}-4}+6\log \mid x+\sqrt{x^{2}-4}\mid +C\]
d) \[\left(x+6\right)\sqrt{x^{2}-4}+2\log \mid x+\sqrt{x^{2}-4}\mid +C\]

Answer: b
Explanation:
q28

9. If \[I=\int\frac{dx}{x^{3}\sqrt{x^{2}-1}}\]
then I is equal to
a) \[\frac{1}{2}\left(\frac{\sqrt{x^{2}-1}}{x}+\tan^{-1}\sqrt{x^{2}-1}\right)+C\]
b) \[\frac{1}{2}\left(\frac{\sqrt{x^{2}-1}}{x^{2}}+\tan^{-1}\sqrt{x^{2}-1}\right)+C\]
c) \[\frac{1}{2}\left(\frac{\sqrt{x^{2}-1}}{x^{3}}+\tan^{-1}\sqrt{x^{2}-1}\right)+C\]
d) \[\frac{1}{2}\left(\frac{\sqrt{x^{2}-1}}{x^{2}}+x\tan^{-1}\sqrt{x^{2}-1}\right)+C\]

Answer: b
Explanation:
q29

10. If \[I=\int\frac{dx}{\left(a^{2}-b^{2}x^{2}\right)^{3/2}}\]
then I is equal to
a) \[\frac{x}{\sqrt{a^{2}-b^{2}x^{2}}}+C\]
b) \[\frac{x}{a^{2}\sqrt{a^{2}-b^{2}x^{2}}}+C\]
c) \[\frac{ax}{\sqrt{a^{2}-b^{2}x^{2}}}+C\]
d) \[\frac{ab}{\sqrt{a^{2}-b^{2}x^{2}}}+C\]

Answer: b
Explanation:
q30
q30a