Mathematics

# Evaluate $\int { \sin ^{ \tfrac{3}{4} }{ x } } \cos { x } dx$.

##### SOLUTION
$\displaystyle\int{{\sin}^{\tfrac{3}{4}}{x}\cos{x}dx}$

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

$=\displaystyle\int{{t}^{\tfrac{3}{4}}dt}$

$=\dfrac{{t}^{\tfrac{3}{4}+1}}{\dfrac{3}{4}+1}+c$

$=\dfrac{{t}^{\tfrac{3+4}{4}}}{\dfrac{3+4}{4}}+c$

$=\dfrac{{t}^{\tfrac{7}{4}}}{\dfrac{7}{4}}+c$

$=\dfrac{4}{7}{t}^{\tfrac{7}{4}}+c$ , where $c$ is the constant of integration

$=\dfrac{4}{7}{\sin}^{\tfrac{7}{4}}{x}+c$, where $t=\sin{x}$

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

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