Mathematics

# Evaluate the following integral:$\int { \cfrac { \sin { 2x } }{ { \left( a+b\cos { 2x } \right) }^{ 2 } } } dx$

##### SOLUTION
Let
$t=a+b\cos{2x}\Rightarrow\,dt=-2b\sin{2x}dx$

$\Rightarrow\,\dfrac{-1}{2b}dt=\sin{2x}dx$

Now,$\displaystyle\int{\dfrac{\sin{2x}}{{\left(a+b\cos{2x}\right)}^{2}}dx}$

$=\dfrac{-1}{2b}\displaystyle\int{\dfrac{dt}{{t}^{2}}}$

$=\dfrac{-1}{2b}\displaystyle\int{{t}^{-2}dt}$

$=\dfrac{-1}{2b}\left[\dfrac{{t}^{-2+1}}{-2+1}\right]+c$

$=\dfrac{1}{2bt}+c$

$=\dfrac{1}{2b\left(a+b\cos{2x}\right)}+c$ where $t=a+b\cos{2x}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
If $\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac { \sin { x } }{ 1+\sin { x+\cos { x } } } }dx=\dfrac{\pi}{a}-\dfrac{1}{2}lnb,\ a,b \in N$ then
• A. $a+b=4$
• B. $a-b=4$
• C. $a-b=6$
• D. $a+b=6$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Hard
Evaluate $\displaystyle \int \frac{e^{x}}{e^{2x}+6e^{x}+5}dx.$
The ans is $\displaystyle =\frac{1}{4}log\left | \frac{e^{x}+1}{e^{x}+ k} \right |+C$
what is k

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle \int _{ \frac { 1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{\displaystyle \frac { x^{ 4 } }{ 1-x^{ 4 } } \ \cos ^{ -1 }{\displaystyle \frac { 2x }{ 1+x^{ 2 } } } dx } =k\int _{ 0 }^{ \frac { 1 }{ \sqrt { 3 } } }{ \displaystyle \frac { x^{ 4 } }{ 1-x^{ 4 } } dx }$,
then $'k'$ is equal to?
• A. $2\pi$
• B. $2$
• C. $1$
• D. $\pi$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle I = \int \frac {\sec^{-1} \sqrt x - cosec^{-1} \sqrt x}{\sec^{-1} \sqrt x + cosec^{-1} \sqrt x} dx$, then I equals
• A. $\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) + x + C$
• B. $\displaystyle (4/\pi) (x \sec^{-1} \sqrt x + \sqrt {x - 1}) - x + C$
• C. none of these
• D. $\displaystyle (4/\pi) (x \sec^{-1} \sqrt x - \sqrt {x - 1}) - x + C$

Evaluate: $\displaystyle\int \tan^{-1}\sqrt{x}dx$.