Question

Integrate the given function w.r.t. respective variable : sin⁴ x

Answer 1

Answer:

$\int{sin^4x}\:dx=\frac{3x}{8}+\frac{sin\:4x}{32}-\frac{sin2x}{4}+C$

Step-by-step explanation:

$\int{sin^4x}\:dx$

$=\int(sin^2x)^2\:dx$

Using

$\boxed{\begin{minipage}{4cm} cos2A=1-2\:sin^2A\\ \\\:\implies\:sin^2A=\frac{1-cos2A}{2} \end{minipage}}$

$=\int(sin^2x)^2\:dx$

$=\int(\frac{1-cos2x}{2})^2\:dx$

$=\int\frac{(1+cos^22x-2\:cos2x)}{4}\:dx$

$=\frac{1}{4}\int{(1+cos^22x-2\:cos2x)}\:dx$

Using

$\boxed{\begin{minipage}{4cm} cos2A=2\:cos^2A-1\\ \\\:\implies\:cos^2A=\frac{1+cos2A}{2} \end{minipage}}$

$=\frac{1}{4}\int{(1+\frac{1+cos4x}{2}-2\:cos2x)}\:dx$

$=\frac{1}{4}\int{\frac{2+1+cos4x-4\:cos2x}{2}}\:dx$

$=\frac{1}{8}\int(3+cos4x-4\:cos2x)\:dx$

$=\frac{1}{8}[3x+\frac{sin\:4x}{4}-4\:\frac{sin2x}{2}]+C$

$=\frac{3x}{8}+\frac{sin\:4x}{32}-\frac{sin2x}{4}+C$

$\implies\:\boxed{\int{sin^4x}\:dx=\frac{3x}{8}+\frac{sin\:4x}{32}-\frac{sin2x}{4}+C}$

Answer 2

Answer:

$\mathbf{\frac{3x}{8}+\frac{sin\:4x}{32}-\frac{sin2x}{4}+C}$

Step-by-step explanation:

In this question

We have to integrate the function

$\int{sin^4x}\:dx$

=$\int(sin^2x)^2\:dx$

Using

$\mathbf{\begin{minipage}{4cm} cos2A=1-2\:sin^2A\\ \\\:\implies\:sin^2A=\frac{1-cos2A}{2} \end{minipage}}$

= $\int(sin^2x)^2\:dx$

= $\int(\frac{1-cos2x}{2})^2\:dx$

= $\int\frac{(1+cos^22x-2\:cos2x)}{4}\:dx$

= $\frac{1}{4}\int{(1+cos^22x-2\:cos2x)}\:dx$

Again Using

$\mathbf{\begin{minipage}{4cm} cos2A=2\:cos^2A-1\\ \\\:\implies\:cos^2A=\frac{1+cos2A}{2} \end{minipage}}$

= $\frac{1}{4}\int{(1+\frac{1+cos4x}{2}-2\:cos2x)}\:dx$

= $\frac{1}{4}\int{\frac{2+1+cos4x-4\:cos2x}{2}}\:dx$

= $\frac{1}{8}\int(3+cos4x-4\:cos2x)\:dx$

= $\frac{1}{8}[3x+\frac{sin\:4x}{4}-4\:\frac{sin2x}{2}]+C$

= $\frac{3x}{8}+\frac{sin\:4x}{32}-\frac{sin2x}{4}+C$

$\implies\:\mathbf{\int{sin^4x}\:dx=\frac{3x}{8}+\frac{sin\:4x}{32}-\frac{sin2x}{4}+C}$