Mathematics

# The value of $\displaystyle\int_{-\pi}^{\pi}\sin^{3}x \cos^{2}x\ dx$ is equal to

$0$

##### SOLUTION
applying, $\displaystyle \int_{a}^{b}+(x)dx = \int_{a}^{b}f(a+b-x)dx...$
$\displaystyle \int_{-\pi }^{\pi } sin^{3}x\,cos^{2}dx = \int_{-\pi }^{\pi } sin^{3}x(1-sin^{2}x)dx$
$\displaystyle\Rightarrow \int_{-\pi }^{\pi }(sin^{3}x-sin^{5}x)dx = I....(1)$
$\displaystyle \Rightarrow I = \int_{-\pi }^{\pi }(sin^{3}(\pi -\pi -x)-sin^{5}x(\pi -\pi -x))dx$
$\displaystyle I = \int_{-\pi }^{\pi } (-sin^{3}x+sin^{5}x)dx....(2)$
$(1)+(2) \Rightarrow 2\pi = 0$
$\boxed {I = 0}$

Its FREE, you're just one step away

Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int \dfrac {4dx}{x^{2}\sqrt {4 - 9x^{2}}}$ is equal to
• A. $\sqrt {4 - 9x^{2}} + C$
• B. $\dfrac {-2}{3} \sqrt {4 - 9x^{2}} + C$
• C. $\dfrac {2}{3} \dfrac {\sqrt {4 - 9x^{2}}}{x} + C$
• D. $-3 \sqrt {4 - 9x^{2}} + C$
• E. $\dfrac {-\sqrt {4 - 9x^{2}}}{x} + C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \overset{-\pi/2}{\underset{-3\pi/2}{\int}} \{ (x + \pi)^3 \} dx =$
• A. $\dfrac{\pi}{2}$
• B. $\dfrac{5\pi^4}{16}$
• C. none of these
• D. $\dfrac{5\pi^4}{8}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate:
$\displaystyle \int e^{tan^{-1}x} \left ( \dfrac{1 + x + x^2}{1 + x^2} \right ) dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve :

$\displaystyle \int_{1}^{2} \dfrac {x}{(x+1)(x+2)}dx$

$(x^2+1) \log x$