Mathematics

# $\displaystyle \int\frac{\sin x.\cos x}{1+\sin^{4}x}d_{X}=$

$\displaystyle \frac{1}{2}\tan^{-1}(\sin^{2}x)+c$

##### SOLUTION
=$\displaystyle \frac {1}{2}\int \frac {\sin 2x}{1+\sin^4x}dx$
=$\displaystyle \frac {1}{2}\int \frac {d \sin^2x}{1+(\sin^2x)^2}=\frac {1}{2} \tan^{-1}(\sin^2x)+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Subjective Medium
$\int_{ - 1}^2 {\sqrt {5x + 6} } dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int \dfrac{1}{3\sin x +4\cos x} dx$ equal, where $\tan \alpha = \dfrac{4}{3}$
• A. $\dfrac{1}{5} ln \left|\tan \left(x+\alpha\right)\right|+K$
• B. $\dfrac{1}{5} ln \left|\tan \left(\dfrac{x}{2} +\dfrac{\pi}{2}- \dfrac{\alpha}{2}\right)\right|+K$
• C. $\dfrac{1}{5} ln \left|\tan \left(\dfrac{x}{2} -\dfrac{\alpha}{2}\right)\right|+K$
• D. $\dfrac{1}{5} ln \left|\tan \left(\dfrac{x}{2} + \dfrac{\alpha}{2}\right)\right|+K$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Solve $\displaystyle \int _{ 0 }^{ 1/2 }{ \dfrac { x\sin ^{ -1 }{ x } }{ \sqrt { 1-{ x }^{ 2 } } } dx= }$
• A. $\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 3 } \pi }{ 12 }$
• B. $\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 2 } \pi }{ 12 }$
• C. $None\ of\ these$
• D. $\dfrac { 1 }{ 2 } +\dfrac { \sqrt { 3 } \pi }{ 12 }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following integral:

$\displaystyle \int { \cfrac { { e }^{ x } }{ { e }^{ 2x }+5{ e }^{ x }+6 } } dx\quad$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$