Mathematics

# Find $\displaystyle \int \dfrac {1}{\sin x\cos^{3}x}dx.$

##### SOLUTION
$\int { \dfrac { 1 }{ \sin { x } \cos ^{ 3 }{ x } } } dx$
$=\int { \dfrac { 1 }{ \dfrac { \sin { x } }{ \cos { x } } .\cos { x } .\cos ^{ 3 }{ x } } } dx$
$=\int { \dfrac { 1 }{ \tan { x } \cos ^{ 4 }{ x } } } dx$
$=\int { \dfrac { \sec ^{ 4 }{ x } }{ \tan { x } } } dx$       $\left( \because \cos x=\dfrac{1}{\sec x}\right)$
Now, substitute $t=\tan x$
$\Rightarrow \dfrac{dt}{dx}=\sec^2 x$
$\Rightarrow dt=\sec^2 xdx$
$=\int { \dfrac { \sec ^{ 2 }{ x } \sec ^{ 2 }{ x } }{ \tan { x } } } dx$
$=\int { \dfrac { \left( 1+\tan ^{ 2 }{ x } \right) \sec ^{ 2 }{ x } }{ \tan { x } } } dx$      $\left( \because \sec^2 x-\tan^x=1 \rightarrow \sec^2 x=1+\tan^2 x\right)$
$=\int { \dfrac { \left( 1+t^2 \right) }{ t } } dx$
$=\int \dfrac{1}{t}dt +\int t dt$
$=ln\;t+\dfrac{t^2}{2}+c$
$=ln \left| \tan^{-1}x\right| +\dfrac{\left( \tan^{-1}x\right)^2}{2}+c$
Hence, the answer is $ln \left| \tan^{-1}x\right| +\dfrac{\left( \tan^{-1}x\right)^2}{2}+c.$

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Subjective Medium
Evaluate the integrals : $\displaystyle \int \dfrac{x^5}{\sqrt{1+x^3}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\displaystyle \int_{0}^{\pi /2}\frac{dx}{1+\tan ^{3}x}$ is
• A. $\displaystyle \frac{\pi }{2}$
• B. $\pi$
• C. none of these
• D. $\displaystyle \frac{\pi }{4}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle\int { \dfrac { 1 }{ { x }^{ 2 }{ \left( { x }^{ 4 }+1 \right) }^{ { 3 }/{ 4 } } } dx }$ is equal to
• A. ${ \left( 1+\dfrac { 1 }{ { x }^{ 4 } } \right) }^{ { 1 }/{ 4 } }+C$
• B. ${ \left( { x }^{ 4 }+1 \right) }^{ { 1 }/{ 4 } }+C$
• C. ${ \left( 1-\dfrac { 1 }{ { x }^{ 4 } } \right) }^{ { 1 }/{ 4 } }+C$
• D. $-{ \left( 1+\dfrac { 1 }{ { x }^{ 4 } } \right) }^{ { 1 }/{ 4 } }+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Evaluate: $\displaystyle \int_{0}^{\pi}x \sin^{3}xdx$
• A. $\displaystyle \frac{\pi}{3}$
• B. $\displaystyle \frac{-2\pi}{3}$
• C. $2 \pi$
• D. $\displaystyle \frac{2\pi}{3}$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$