Trigonometric integral evaluation: $\int 4 \sin^4 x \cos^3 x \,dx$

Question

Evaluate the following integral $$\int 4 \sin^4 x \cos^3 x \,dx$$

I can do simple integration problems, but problems like this seem to stump me, I created this problem so I could solve and compare it to another similar problem for my study guide, where should I start?

Answer 1

Hint: If you use the Pythagorean identity $\cos^2(x) = 1 - \sin^2(x)$, then you can reduce the integral to

$$\int 4 \sin^4(x)(1-\sin^2(x))\cos(x)\, dx.$$

Can you see what to do from here?

Answer 2

\begin{align} 4\int \sin^4{x}\cos^3{x}dx &=4\int \sin^4{x}(1-\sin^2{x})\cos{x} \ dx\\ &=4\int \sin^4{x}-\sin^6{x} \ d(\sin{x})\\ &=\frac{4}{5}\sin^5{x}-\frac{4}{7}\sin^7{x}+c \end{align}

Answer 3

For "trigonometric powers" integrals where the two functions are "linked" by the Pythagorean Identity, you want to try to arrange for one of the factors, which is a power of one trig function, to be written in terms of an even power of the other trig function. So for powers of sine and cosine in the integral $ \ \int \ \sin^m x \ \cos^n x \ \ dx \ $ , we have

for $ \ m \ $ odd and $ \ n \ $ even -- extract a factor of $ \ \sin x \ $ to place with the differential $ \ dx \ $ , writing $ \ \int \ \sin^{m-1} x \ \cos^n x \ \ ( \sin x \ dx) \ $ , then make the substitution $ \ du \ = \ \sin \ x \ dx \ \Rightarrow \ u \ = \ -\cos \ x \ $ ; with $ \ m - 1 \ = \ 2p \ $ , we have

$$ \ \int \ \sin^{2p} x \ \cos^n x \ \ ( \sin x \ dx) \ \ \Rightarrow \ \ \int \ (1 - \cos^2 \ x)^p \ \cos^n x \ \ ( \sin x \ dx) $$

$$ \Rightarrow \ \ \int \ (1 - u^2 )^p \ (-u)^n \ \ du \ \ ; $$

for $ \ m \ $ even and $ \ n \ $ odd -- extract a factor of $ \ \cos x \ $ , write $ \ \int \ \sin^m x \ \cos^{n-1} x \ \ ( \cos x \ dx) \ $ , $ \ du \ = \ \cos \ x \ dx \ \Rightarrow \ u \ = \ \sin \ x \ $ ; with $ \ n - 1 \ = \ 2q \ $ , we have

$$ \ \int \ \sin^m x \ \cos^{2q} x \ \ ( \cos x \ dx) \ \ \Rightarrow \ \ \int \ \sin^m x \ (1 - \sin^2 \ x)^q \ \ ( \cos x \ dx) $$

$$ \Rightarrow \ \ \int \ u^m \ (1 - u^2 )^q \ \ du \ \ ; $$

for $ \ m \ $ and $ \ n \ $ both odd -- use either of the above;

for $ \ m \ $ and $ \ n \ $ both even-- we use the "sine/cosine-squared identities", $ \ \sin^2 \ x \ = \ \frac{1}{2} \ (1 \ - \ \cos \ 2x) \ $ $ \ \cos^2 \ x \ = \ \frac{1}{2} \ (1 \ + \ \cos \ 2x) \ $ ; with $ \ m \ = \ 2p \ $ and $ \ n \ = \ 2q \ $ , we have

$$ \ \int \ \sin^{2p} x \ \cos^{2q} x \ \ dx \ \ \Rightarrow \ \ \int \ \left[ \ \frac{1}{2} \ (1 \ - \ \cos \ 2x) \ \right]^p \ \left[ \ \frac{1}{2} \ (1 \ + \ \cos \ 2x) \ \right]^q \ \ dx \ \ . $$

[Special case: if $ \ m \ = \ n \ $ , odd or even , we can just use

$$ \ \int \ \sin^m x \ \cos^m x \ \ dx \ \ \Rightarrow \ \ \int \ ( \ \sin \ x \ \cos \ x \ )^m \ \ dx \ = \ \int \ \left( \ \frac{1}{2} \ \sin \ 2x \ \right)^m \ \ dx \ \ . \ ] $$

You will unavoidably have some sort of binomial expansion to carry out if the smaller power is larger than three. There are analogous methods for powers of tangent and secant, and for powers of cotangent and cosecant.