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# 8.3: Powers of sine and cosine

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Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious, but the technique is straightforward. Some examples will suffice to explain the approach.

Example 8.2.1

Evaluate

$\int \sin^5 x\,dx. \nonumber$

Solution

Rewrite the function:

$\int \sin^5 x\,dx=\int \sin x \sin^4 x\,dx= \int \sin x (\sin^2 x)^2\,dx= \int \sin x (1-\cos^2 x)^2\,dx. \nonumber$

Now use $$u=\cos x$$, $$du=-\sin x\,dx$$:

\eqalign{ \int \sin x (1-\cos^2 x)^2\,dx&=\int -(1-u^2)^2\,du\cr &=\int -(1-2u^2+u^4)\,du\cr &=-u+{2\over3}u^3-{1\over5}u^5+C\cr &=-\cos x+{2\over3}\cos^3 x-{1\over5}\cos^5x+C.\cr } \nonumber

Example 8.2.2

Evaluate

$\int \sin^6 x\,dx. \nonumber$

Solution

Use $$\sin^2x =(1-\cos(2x))/2$$ to rewrite the function:

\eqalign{ \int \sin^6 x\,dx=\int (\sin^2 x)^3\,dx&= \int {(1-\cos 2x)^3\over 8}\,dx\cr &={1\over 8}\int 1-3\cos 2x+3\cos^2 2x-\cos^3 2x\,dx.\cr} \nonumber

Now we have four integrals to evaluate:

$\int 1\,dx=x\nonumber$

and

$\int -3\cos 2x\,dx = -{3\over 2}\sin 2x\nonumber$

are easy. The $$\cos^3 2x$$ integral is like the previous example:

\eqalign{ \int -\cos^3 2x\,dx&=\int -\cos 2x\cos^2 2x\,dx\cr &=\int -\cos 2x(1-\sin^2 2x)\,dx\cr &=\int -{1\over 2}(1-u^2)\,du\cr &=-{1\over 2}\left(u-{u^3\over 3}\right)\cr &=-{1\over 2}\left(\sin 2x-{\sin^3 2x\over 3}\right).}

And finally we use another trigonometric identity, $$\cos^2x=(1+\cos(2x))/2$$:

$\int 3\cos^2 2x\,dx=3\int {1+\cos 4x\over 2}\,dx= {3\over 2}\left(x+{\sin 4x\over 4}\right).\nonumber$

So at long last we get

$\int \sin^6 x\,dx = {x\over8} -{3\over 16}\sin 2x -{1\over 16}\left(\sin 2x-{\sin^3 2x\over 3}\right) +{3\over 16}\left(x+{\sin 4x\over 4}\right)+C. \nonumber$

Example 8.2.3

Evaluate

$\int\! \sin^2x\cos^2x\,dx.\nonumber$

Solution

Use the formulas $$\sin^2x =(1-\cos(2x))/2$$ and $$\cos^2x =(1+\cos(2x))/2$$ to get:

$\int \sin^2x\cos^2x\,dx=\int {1-\cos(2x)\over2}\cdot {1+\cos(2x)\over2}\,dx. \nonumber$

The remainder is left as an exercise.

## Contributors

• Integrated by Justin Marshall.