Mathematics

# Evaluate:$\int { \sin ^{ 3 }{ x } .\cos ^{ 3 }{ x } } dx$

##### SOLUTION
$\displaystyle \int sin^3x cos^3 x dx$

$\displaystyle \Rightarrow \int (1-cos^2 x) sinx cos^3x dx$

$\displaystyle \int sin^3x cosx (1-sin^2 x)dx$

Let $sin x = m$

Differentiating w.r.t. $x$, we get,

$cosx \,dx = dm$

$\displaystyle \Rightarrow \int m^3 (1-m^2)dm$

$\displaystyle \Rightarrow \int m^{3} dm - \int m^5 dm$

$\displaystyle \Rightarrow \frac{m^4}{4}-\frac{m^6}{6}+c$

Back substituting the value of $m$

$\displaystyle \Rightarrow \frac{sin^4x}{4}-\frac{sin^6x}{6}+c$

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

#### Realted Questions

Q1 Subjective Medium
Integrate $\int {\sqrt {1 - {t^2}} } dt$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\int {\frac{{dx}}{{\sqrt {{x^2} - 9} + \sqrt {{x^2} - 4} }} = } ?$
• A. $- \frac{1}{5}\left[ {\frac{x}{2}\left( {\sqrt {{x^2} - 9} - \sqrt {{x^2} - 4} } \right) + \log \left[ {\frac{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^2}}}{{\left( {x + \sqrt {{x^2} - 4} } \right)}}} \right]} \right]$
• B. $\frac{1}{5}\left[ {\frac{x}{2}\left( {\sqrt {{x^2} - 9} + \sqrt {{x^2} - 4} } \right) + \log \left[ {\frac{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^2}}}{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^{9/2}}}}} \right]} \right]$
• C. None of these
• D. $\frac{1}{5}\left[ {\frac{x}{2}\left( {\sqrt {{x^2} - 9} - \sqrt {{x^2} - 4} } \right) + \log \left[ {\frac{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^2}}}{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^{9/2}}}}} \right]} \right]$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Integrate $\int_{0}^{1}\dfrac{\sqrt x}{1+x^3}$
• A. $\dfrac {\pi}{4}$
• B. $\dfrac {\pi}{3}$
• C. $\dfrac {\pi}{2}$
• D. $\dfrac {\pi}{6}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle \int\sqrt{\frac{x-1}{2x-3}}dx=$
• A. $sin^{-1}x+\sqrt{1-x^{2}}+c$
• B. $sin^{-1}x-\sqrt{1-x^{2}}+c$
• C. $2sin^{-1}x-\sqrt{1-x^{2}}+c$
• D. $\dfrac{1}{2}\displaystyle \sqrt{2x^{2}-5x+3}+\frac{1}{4}\cosh^{-1}(4x-5)+c$

$\int { \cfrac { f'(x) }{ f(x) } dx } =\log { [f(x)] } +c$