Mathematics

# $\displaystyle \int \dfrac{\sin^{2} x \cos^{2} x}{(\sin^{2}x+\cos^{3}x\sin^{2}x+\sin^{3}x\cos^{2}x+\cos^{5}x)^{2}} dx$ is equal to:

##### SOLUTION
$\int { \cfrac { \sin ^{ 2 }{ x } .\cos ^{ 3 }{ x } }{ { \left( \sin ^{ 2 }{ x } +\cos ^{ 3 }{ x } .\sin ^{ 2 }{ x } +\sin ^{ 3 }{ x } \cos ^{ 2 }{ x } +\cos ^{ 5 }{ x } \right) }^{ 2 } } dx }$
$=\int { \cfrac { \sin ^{ 2 }{ x } .\cos ^{ 2 }{ x } }{ { \left\{ \left[ \sin ^{ 2 }{ x } \left[ \sin ^{ 3 }{ x } .\cos ^{ 3 }{ x } \right] \right] +\cos ^{ 2 }{ x } \left[ \sin ^{ 3 }{ x } +\cos ^{ 3 }{ x } \right] \right\} }^{ 2 } } dx }$
$\int { \cfrac { \sin ^{ 2 }{ x } .\cos ^{ 2 }{ x } }{ { \left\{ \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \right) \left[ \sin ^{ 3 }{ x } +\cos ^{ 3 }{ x } \right] \right\} }^{ 2 } } dx }$
$=\int { \cfrac { \sin ^{ 2 }{ x } .\cos ^{ 2 }{ x } }{ { \left( \sin ^{ 3 }{ x } +\cos ^{ 3 }{ x } \right) }^{ 2 } } dx }$
Divide $\cos ^{ 3 }{ x }$ in numerator and denominator
$=\int { \cfrac { \sec ^{ 3 }{ x } .\tan ^{ 2 }{ x } }{ { \left( \tan ^{ 3 }{ x } +1 \right) }^{ 2 } } dx }$
Let $1+\tan ^{ 3 }{ x } =t$
$3\tan ^{ 2 }{ x } \sec ^{ 2 }{ x } dx=dt$
$\therefore \int { \cfrac { 1 }{ 3 } \cfrac { dt }{ { t }^{ 2 } } } =\cfrac { -1 }{ 3 } \cfrac { 1 }{ t } +c$
$=\cfrac { 1 }{ 3\left( \tan ^{ 3 }{ x } +1 \right) } +c$

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

#### Realted Questions

Q1 Single Correct Hard
If $f(x)=\begin{vmatrix}\cos{x}&e^{\displaystyle x^2}&\displaystyle 2x\cos^2{\frac{x}{2}}\\x^2&\sec{x}&\sin{x}+x^3\\1&2&x+\tan{x}\end{vmatrix}$, then the value of $\displaystyle\int_{\displaystyle-\frac{\pi}{2}}^{\displaystyle\frac{\pi}{2}}{(x^2+1)(f(x)+f^{\prime\prime}(x))dx}$, is equal to
• A. 1
• B. $-1$
• C. 2
• D. none of these

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int_{0}^{\pi /2n}\dfrac{dx}{1+cot ^{n}nx}dx=$
• A. $\displaystyle \frac{\pi}{6n}$
• B. $\displaystyle \frac{\pi}{8n}$
• C. $\displaystyle \frac{\pi}{12n}$
• D. $\displaystyle \frac{\pi}{4n}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\displaystyle I = \int e^x \frac {x^3 + 3x^2 + 4}{(x + 1)^3} dx$, then I equals
• A. $\displaystyle e^x \left ( \frac {x^2 - x + 1}{(x + 1)^2} \right ) + C$
• B. $\displaystyle e^{-x} \left ( \frac {x^2 - x + 1}{(x - 1)^2} \right ) + C$
• C. $\displaystyle e^x \left ( \frac {x^2 + 2x - 2}{(x + 1)^2} \right ) + C$
• D. none of these

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate $\displaystyle\int^{\pi/3}_0\cos^3xdx$.

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$