Mathematics

# Evaluate :$\displaystyle{{\int}_{0}}^{\pi /2} \dfrac{\sin x}{1+cos^2x}dx$

##### SOLUTION
$I={{\int}_{0}}^{\pi /2} \dfrac{\sin x }{1+cos^2x}dx$

Let $\cos x =t$
$-\sin x dx=dt$
Now, $x=0 \implies t=1$ and $x=\dfrac{\pi}{2} \implies t=0$

Therefore,
$I={{\int}_{1}}^0 \dfrac{\sin x}{1+t^2}(\dfrac{-dt}{\sin x})$

$I=-{{\int}_{1}}^0 \dfrac{dt}{1+t^2}$

$I=-{[tan^{-1}t]_{1}}^0$

$I=-[tan^{-1}0-tan^{-1}1]=-[0-\dfrac{\pi}{4}]=\dfrac{\pi}{4}$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Subjective Medium
$\displaystyle \int \dfrac{x+3}{(x+1)^{4}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\int { { ({ x }^{ 2 }+5) }^{ 3 } } dx$
• A. $\cfrac { { x }^{ 6 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$
• B. $\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }-25{ x }^{ 3 }+125x+C$
• C. $\cfrac { { x }^{ 7 } }{ 7 } +4{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$
• D. $\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
solve$\int(1+x)^{2}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
The value of $\displaystyle\int \dfrac {\sin 2x}{\sin^{4}x + \cos^{4}x} dx$ is
• A. $\tan^{-1} (\cot^{2}x) + C$
• B. $\tan^{-1} (\sin 2x) + C$
• C. $\tan^{-1} (\tan^{2} x) + C$
• D. $-\tan^{-1} (\cos 2x) + C$

Find $\displaystyle \int_{0}^{\pi/2} \sin x. \sin 2x\ dx$