Evaluate the integral by making the given substitution. (Use C for the constant of integration.)? 
sin5(θ) cos(θ) dθ, u= sin(θ)

Solution:

Given, \(\int sin^{5}(\theta )cos(\theta )\, d\theta\)

Consider u = sin(θ)

We have to evaluate the integral by using C for the constant of integration.

Given, u = sin(θ)

du = cos(θ) dθ

So, dθ = 1/cos(θ) du

Now, \(\int sin^{5}(\theta )cos(\theta )\, d\theta = \int u^{5}cos(\theta )\frac{1}{cos(\theta )}du\)

= \(\int u^{5}du\)

We know, \(\int y^{n}\, dy = \frac{y^{n+1}}{n+1}\)

So, \(\int u^{5}\, du=\frac{u^{5+1}}{5+1}+C\)

= \(\frac{u^{6}}{6}+C\)

Now, substitute u = sin(θ)

 =\(\frac{sin^{6}(\theta )}{6}+C\)

Therefore, \(\int sin^{5}(\theta )cos(\theta )\, d\theta =\frac{sin^{6}(\theta )}{6}+C\)

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Evaluate the integral by making the given substitution. (Use C for the constant of integration.)? 
sin5(θ) cos(θ) dθ, u = sin(θ)

Summary:

 On evaluating the integral by making the given substitution. (Use C for the constant of integration.)

sin⁵(θ) cos(θ) dθ, u= sin(θ) is \(\frac{sin^{6}(\theta )}{6}+C\).