Trigonometric integrals

Integration: Integration techniques

Trigonometric integrals

An integral that only contains trigonometric functions is called a trigonometric integral. These integrals can often be calculated using the substitution method. The challenge is knowing which substitution to use. In many cases, we can use the following rule.

Substitution for trigonometric integrals
Let #\blue{m} \geq 0# and #\orange{n} \geq 0# be integers. Then the integral #\int \sin(x)^{\blue{m}} \cdot \cos(x)^{\orange{n}} \; \dd x# can be solved using the substitution method with the following substitutions.

If #\blue{m}# is odd, then we use #\green{h(x)}=\green{\cos(x)}#.
If #\orange{n}# is odd, then we use #\green{h(x)}=\green{\sin(x)}#.
If #\blue{m}# and #\orange{n}# are even, then we use #\green{h(x)}=\green{2 \cdot x}#.

Note that if #\blue{m}# and #\orange{n}# are both odd, we can choose whether to use the substitution #\green{h(x)}=\green{\cos(x)}# or #\green{h(x)}=\green{\sin(x)}#.

Explanation of substitution
This means that we can write the integral #\int \sin(x)^{\blue{m}} \cdot \cos(x)^{\orange{n}} \; \dd x# as #\int g(\green{h(x)}) \; \dd \green{h(x)}# with the #h(x)# from the box and a suitable #g(x)#.

Definite integral
These substitutions also work for a definite integral instead of an indefinite integral.

Case #\blue{m}# is odd
In this case, we know that #\blue{m}# is odd and therefore #\blue{m}-1# is even. Here we will show why the substitution #\green{h(x)}=\green{\cos(x)}# is useful in this case. Note that \[\dd \green{h(x)} = \dd \, \green{\cos(x)} = {-\sin(x)} \, \dd x\] We first rewrite the integral before applying the substitution. \[\begin{array}{rcl} \displaystyle\int \sin(x)^m\cdot \green{\cos(x)}^n \, \dd x &=& \displaystyle \int \sin(x)^{\blue{m}-1} \cdot \sin(x) \cdot \green{\cos(x)}^{\orange{n}} \, \dd x \\
&&\quad \blue{\text{used }\sin(x)^{\blue{m}}=\sin(x)^{\blue{m}-1} \cdot \sin(x)} \\
&=& \displaystyle \int \left(1-\green{\cos(x)}^2\right)^{\frac{\blue{m}-1}{2}} \cdot {\sin(x)} \cdot \green{\cos(x)}^{\orange{n}} \, \dd x \\
&&\quad \blue{\text{formula }1=\sin(x)^2+\cos(x)^2 \text{ applied}} \\
&=& \displaystyle \int -\left(1-\green{\cos(x)}^2\right)^{\frac{\blue{m}-1}{2}} \cdot \green{\cos(x)}^{\orange{n}} \cdot \left(-\sin(x)\right) \, \dd x \\
&&\quad \blue{\text{written differently by using }-1 \cdot -1 =1} \\
&=& \displaystyle \int -\left(1-\green{\cos(x)}^2\right)^{\frac{\blue{m}-1}{2}} \cdot \green{\cos(x)}^{\orange{n}} \, \dd \green{\cos(x)} \\
&&\quad \blue{\text{used that }\frac{\dd}{\dd x}{\cos(x)}=-\sin(x)} \\
&=& \displaystyle \int - \left(1-\green{u}^2\right)^{\frac{\blue{m}-1}{2}} \cdot \green{u}^\orange{n} \, \dd \green{u} \\
&&\quad \blue{\text{used substitution }{u}={\cos(x)}} \end{array}\] The integral we end up with can then be calculated.

Case #\orange{n}# is odd
In this case, we know that #\orange{n}# is odd and therefore #\orange{n}-1# is even. Here we will show why the substitution #\green{h(x)}=\green{\sin(x)}# is useful in this case. Note that \[\dd \green{h(x)} = \dd \, \green{\sin(x)} = {\cos(x)} \, \dd x\]Before applying the substitution we first rewrite the integral. \[\begin{array}{rcl} \displaystyle\int \green{\sin(x)}^\blue{m}\cdot \cos(x)^\orange{n} \, \dd x &=& \displaystyle \int \green{\sin(x)}^\blue{m}\cdot \cos(x)^{\orange{n}-1} \cdot \cos(x) \, \dd x \\
&&\quad \blue{\text{used }\cos(x)^{n}=\cos(x)^{{n}-1} \cdot \cos(x)} \\
&=& \displaystyle \int \green{\sin(x)}^\blue{m}\cdot \left(1-\green{\sin(x)}^2 \right)^{\frac{\orange{n}-1}{2}} \cdot {\cos(x)} \, \dd x \\
&&\quad \blue{\text{formula }1=\sin(x)^2+\cos(x)^2 \text{ applied}} \\
&=& \displaystyle \int \green{\sin(x)}^\blue{m}\cdot \left(1-\green{\sin(x)}^2 \right)^{\frac{\orange{n}-1}{2}} \, \dd \green{\sin(x)} \\
&&\quad \blue{\text{used that }\frac{\dd}{\dd x}{\sin(x)}=\cos(x)} \\
&=& \displaystyle \int \green{u}^\blue{m}\cdot \left(1-\green{u}^2\right)^{\frac{n-1}{2}} \, \dd \green{u} \\
&&\quad \ \blue{\text{used substitution }{u}={\sin(x)}} \end{array}\] The integral we end up with can then be calculated.

Case #\blue{m}# and #\orange{n}# are even
In this case, we know that both #\blue{m}# and #\orange{n}# are even. Here we will show why the substitution #\green{h(x)}=\green{2 \cdot x}# is useful in this case. Note that \[\dd \green{h(x)} = \dd \, (\green{2 \cdot x}) = {2} \, \dd x\] We will use the trigonometric formulas \[\cos(x)^2 = \frac{1+\cos(2 \cdot x)}{2} \quad \text{and} \quad \sin(x)^2=\frac{1-\cos(2\cdot x)}{2}\] We first rewrite the integral before applying the substitution \[\begin{array}{rcl} \displaystyle\int \sin(x)^\blue{m}\cdot \cos(x)^\orange{n} \, \dd x &=& \displaystyle \int \left(\sin(x)^2\right)^\frac{\blue{m}}{2} \cdot \left(\cos(x)^2\right)^\frac{\orange{n}}{2} \, \dd x \\
&&\quad \blue{\text{written differently}} \\
&=&\displaystyle \int \left(\frac{1-\cos(2 \cdot x)}{2}\right)^\frac{\blue{m}}{2} \cdot \left(\frac{1+\cos(2\cdot x)}{2}\right)^\frac{\orange{n}}{2} \, \dd x \\
&&\quad \blue{\text{trigonometric formulas applied}} \\
&=& \displaystyle \int \frac{1}{2^{\frac{\blue{m}+\orange{n}}{2}}} \cdot (1-\cos(2 \cdot x))^\frac{\blue{m}}{2} \cdot (1+\cos(2 \cdot x))^\frac{\orange{n}}{2} \, \dd x \\
&&\quad \blue{\text{written differently}} \\
&=& \displaystyle \int \frac{1}{2^{\frac{\blue{m}+\orange{n}}{2}+1}} \cdot (1-\cos(\green{2 \cdot x}))^\frac{\blue{m}}{2} \cdot (1+\cos(\green{2 \cdot x}))^\frac{\orange{n}}{2} \cdot {2} \, \dd x \\
&&\quad \blue{\text{multiplied by }1=\frac{1}{2}\cdot 2} \\
&=& \displaystyle \int \frac{1}{2^{\frac{\blue{m}+\orange{n}}{2}+1}} \cdot (1-\cos(\green{2 \cdot x}))^\frac{\blue{m}}{2} \cdot (1+\cos(\green{2 \cdot x}))^\frac{\orange{n}}{2} \, \dd \green{2 \cdot x} \\
&&\quad \blue{\text{used that }\frac{\dd}{\dd x}{2 \cdot x}=2} \\
&=& \displaystyle \frac{1}{2^{\frac{\blue{m}+\orange{n}}{2}+1}} \cdot\int (1-\cos(\green{u}))^\frac{\blue{m}}{2} \cdot (1+\cos(\green{u}))^\frac{\orange{n}}{2} \, \dd \green{u} \\
&&\quad \blue{\text{used substitution }u={2 \cdot x}} \end{array}\] We can now calculate this integral or use the statement again where the choice of substitution is dependent on whether #\frac{\blue{m}}{2}# and #\frac{\orange{n}}{2}# are even or odd.

These substitutions are not always straightforward to apply, as we saw in the tab "Proof". Often, we need one or more of the following trigonometric identities to rewrite the integral.

Trigonometric identities
\[\sin(x)^2 + \cos(x)^2 = 1 \]

\[\cos(x)^2 = \frac{\cos(2 \cdot x)+1}{2}\]

\[\sin(x)^2 = \frac{1-\cos(2 \cdot x)}{2}\]

What is it used for?
We use these identities to find the right #g(x)# so that we can write the integral #\int \sin(x)^{\blue{m}} \cdot \cos(x)^{\orange{n}} \; \dd x# as #\int g(\green{h(x)}) \; \dd \green{h(x)}#.

Determine #\int \sin ^3\left(x\right) \,\dd x#.

Use #C# to denote the constant of integration.

#\int \sin ^3\left(x\right) \,\dd x=# #{{\cos ^3\left(x\right)}\over{3}}-\cos \left(x\right) + C#

We apply the substitution method with #g(x)=x^2-1# and #h(x)=\cos \left(x\right)#, because in that case #g(h(x)) \cdot h'(x)=\sin ^3\left(x\right)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \sin ^3\left(x\right) \,\dd x&=& \displaystyle \int \left(\cos ^2\left(x\right)-1\right) \cdot -\sin \left(x\right) \, \dd x \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(x)) \cdot h'(x) \, \dd x \text{ with } h'(x)=-\sin \left(x\right)} \\ &&\phantom{xxx}\blue{\text{used the trigonometric rule }\sin^2(x)=1-\cos^2(x)} \\ &=& \displaystyle \int \left(\cos ^2\left(x\right)-1 \right) \, \dd(\cos \left(x\right)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(x)=\dd (h(x))} \\ &=& \displaystyle \int u^2-1 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos \left(x\right)=u} \\ &=& \displaystyle {{u^3}\over{3}}-u +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\cos ^3\left(x\right)}\over{3}}-\cos \left(x\right) +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos \left(x\right)}
\end{array}\]

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