11.5 Integration of trigonometric functions

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  We know that the derivative of $\cos(x)$ is $-\sin(x)$, so we can replace $\sin(x)$ by $-\frac{du}{dx}$, which we then integrate w.r.t $x$, so the overall integral is integrated w.r.t $u$.

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  As we can replace $\cos(x)$ with $u$ and $\cos(x)$ with $u^{3}$, this simplifies the expression considerably.

Proceeding,

Which we can split into partial fractions:

From where we can come up with some simultaneous equations

From this we can deduce that $A=1$, $B=-1$ and $C=0$. Overall, then, we have

Note that another way to do the last four steps is

which is possibly nicer.

Note that while a lot of these integrals are in the formula sheet ⁶⁶ 6 At least, in the OCR A formula booklet, it is given that $\frac{d}{dx}\left[\cot(x)\right]=-\csc^{2}(x)$ and $\frac{d}{dx}\left[\csc(x)\right]=-\csc(x)\cot(x)$ some of them are not (i.e. we have no idea what the integral of $\cot^{2}(x)$ is at present). To get around this, we can use the Pythagorean identity to reduce this problem (the answer to which we don’t know) to a problem which we do know the answer to! Take $\cos^{2}(x)+\sin^{2}(x)=1$, divide through by $\sin^{2}(x)$ and obtain that $\cot^{2}(x)+1=\csc^{2}(x)\implies\cot^{2}(x)=\csc^{2}(x)-1$.

Then,

A way to think about this which isn’t particularly elegant, but is usually quite effective, is to think about all the possible identities which could work, and try each one.

e.g. thinking about identities which involve $\cos(x)$, we have ⁷⁷ 7 These are discussed in the trigonometry section of this document.

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  $\cos^{2}(x)+\sin^{2}(x)\equiv 1$, which doesn’t help here because there’s no $\cos^{2}(x)$ anywhere.

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  $\cos(a+b)\equiv\cos(a)\cos(b)-\sin(a)\sin(b)$ - also no help

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  $\cos(2x)\equiv\left[\cos(x)\right]^{2}-\left[\sin(x)\right]^{2}$, and its two other forms (rearranging using the first bullet point), $\cos(2x)\equiv 1-2\left[\sin(x)\right]^{2}$ and $\cos(2x)\equiv 2\left[\cos(x)\right]^{2}-1$.

The last identity looks quite useful, because we have a $\cos(2x)$, and a $1$, and we can rewrite it in the form

The $1-\cos(2x)$ looks quite a lot, but not exactly, like our integral, but we can fix that by simply replacing ⁸⁸ 8 This is fine, because when we derived the double-angle formula from the addition formula we set $a=x$ and $b=x$ in the equation $\cos(a+b)\equiv\cos(a)\cos(b)-\sin(a)\sin(b)$, but we could have just as well have set $a=\frac{x}{2}$ and $b=\frac{x}{2}$, which would give $\cos(x)\equiv\left[\cos\left(\frac{x}{2}\right)\right]^{2}-\left[\sin\left(\frac{x}{2}\right)\right]^{2}$ which we can then rearrange to give the result we need. $x$ with $\frac{x}{2}$, giving

Applying this to our integral, we have

The last part we can do by inspection (as the derivative of $-\cos(x)$ is equal to $\sin(x)$, by the chain rule, the derivative of $-\cos\left(\frac{x}{2}\right)$ is equal to $-\frac{1}{2}\cos(\frac{x}{2})$, and therefore the integral of $\sin\left(\frac{x}{2}\right)$ with respect to $x$ is just $-2\cos\left(\frac{x}{2}\right)$).

Therefore, we have