Mathematics

# Evaluate the following integrals:$\int { \sin ^{ 3 }{ x } \cos ^{ 5 }{ x } } dx$

##### SOLUTION
$\displaystyle\int{{\sin}^{3}{x}{\cos}^{5}{x}dx}$

$=\displaystyle\int{{\sin}^{3}{x}{\cos}^{2}{x}{\cos}^{3}{x}dx}$

$=\displaystyle\int{{\sin}^{3}{x}\left(1-{\sin}^{2}{x}\right){\cos}^{3}{x}dx}$

$=\displaystyle\int{\left({\sin}^{3}{x}-{\sin}^{5}{x}\right){\cos}^{3}{x}dx}$

$=\displaystyle\int{\left({\sin}^{3}{x}-{\sin}^{5}{x}\right){\cos}^{2}{x}\cos{x}dx}$

$=\displaystyle\int{\left({\sin}^{3}{x}-{\sin}^{5}{x}\right)\left(1-{\sin}^{2}{x}\right)\cos{x}dx}$

$=\displaystyle\int{\left({\sin}^{3}{x}-2{\sin}^{5}{x}+{\sin}^{7}{x}\right)\cos{x}dx}$

$=\displaystyle\int{{\sin}^{3}{x}\cos{x}dx}-2\displaystyle\int{{\sin}^{5}{x}\cos{x}dx}+\displaystyle\int{{\sin}^{7}{x}\cos{x}dx}$

Let

$t=\sin{x}\Rightarrow\,dt=\cos{x}dx$

$=\displaystyle\int{{t}^{3}dt}-2\displaystyle\int{{t}^{5}dt}+\displaystyle\int{{t}^{7}dt}$

$=\dfrac{{t}^{4}}{4}-2\times\dfrac{{t}^{6}}{6}+\dfrac{{t}^{8}}{8}+c$

$=\dfrac{{\sin}^{4}{x}}{4}-\dfrac{{\sin}^{6}{x}}{3}+\dfrac{{\sin}^{8}{x}}{8}+c$ where

$t=\sin{x}$

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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 Medium
Resolve $\displaystyle \frac{1-2x}{3+2x-x^2}$ into partial fractions.
• A. $\displaystyle \frac{5}{4(3-x)}+\frac{3}{4(1+x)}$
• B. $\displaystyle -\frac{5}{2(3-x)}+\frac{3}{4(1+x)}$
• C. $\displaystyle -\frac{5}{4(3-x)}+\frac{3}{2(1+x)}$
• D. $\displaystyle -\frac{5}{4(3-x)}+\frac{3}{4(1+x)}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\displaystyle \int \sqrt{x + \sqrt{x^2 + 2}}$ $dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate:
$\displaystyle\int^2_1\dfrac{dx}{x(x^4+1)}$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate the given integral.
$\displaystyle \int { \cfrac { \log { \left( \log { x } \right) } }{ x } } dx$

Prove that $\displaystyle\int_0^{\pi/2}$ $ln(\sin x)dx=\displaystyle\int_0^{\pi/2}ln(cos x)dx=\int_0^{\pi/2}\,\,ln(sin2x)dx=-\dfrac{\pi}{2}.ln 2$.