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 $$m$$

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

View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

Realted Questions

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

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$ \frac{1}{5}\left[ {\frac{x}{2}\left( {\sqrt {{x^2} - 9} - \sqrt {{x^2} - 4} } \right) + \log \left[ {\frac{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^2}}}{{{{\left( {x + \sqrt {{x^2} - 4} } \right)}^{9/2}}}}} \right]} \right]$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$\dfrac {\pi}{6}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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. $$\dfrac{1}{2}\displaystyle \sqrt{2x^{2}-5x+3}+\frac{1}{4}\cosh^{-1}(4x-5)+c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Medium
$$\int { \cfrac { f'(x) }{ f(x) } dx } =\log { [f(x)] } +c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer