Integration: Integration techniques
Trigonometric integrals
Using the substitution method, we can also solve trigonometric integrals. We often use the following trigonometric rules of calculation here.
\[\sin^2(x) + \cos^2(x) = 1 \]
\[\cos^2(x) = \frac{\cos(2x)+1}{2}\]
\[\sin^2(x) = \frac{1-\cos(2x)}{2}\]
#\int \cos ^9\left(x\right)\cdot \sin \left(x\right) \,\dd x=# #-{{\cos ^{10}\left(x\right)}\over{10}} + C#
We apply the substitution method with #g(x)=-x^9# and #h(x)=\cos \left(x\right)#, because in that case #g(h(x)) \cdot h'(x)=\cos ^9\left(x\right)\cdot \sin \left(x\right)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos ^9\left(x\right)\cdot \sin \left(x\right) \,\dd x&=& \displaystyle \int -\cos ^9\left(x\right) \cdot -\sin \left(x\right) \, \dd x \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(x)) \cdot h'(x) \, \dd x \text{ with } h'(x)=-\sin \left(x\right)} \\ &=& \displaystyle \int \left(-\cos ^9\left(x\right) \right) \, \dd(\cos \left(x\right)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(x)=\dd (h(x))} \\ &=& \displaystyle \int -u^9 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos \left(x\right)=u} \\ &=& \displaystyle -{{u^{10}}\over{10}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos ^{10}\left(x\right)}\over{10}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos \left(x\right)}
\end{array}\]
We apply the substitution method with #g(x)=-x^9# and #h(x)=\cos \left(x\right)#, because in that case #g(h(x)) \cdot h'(x)=\cos ^9\left(x\right)\cdot \sin \left(x\right)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos ^9\left(x\right)\cdot \sin \left(x\right) \,\dd x&=& \displaystyle \int -\cos ^9\left(x\right) \cdot -\sin \left(x\right) \, \dd x \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(x)) \cdot h'(x) \, \dd x \text{ with } h'(x)=-\sin \left(x\right)} \\ &=& \displaystyle \int \left(-\cos ^9\left(x\right) \right) \, \dd(\cos \left(x\right)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(x)=\dd (h(x))} \\ &=& \displaystyle \int -u^9 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos \left(x\right)=u} \\ &=& \displaystyle -{{u^{10}}\over{10}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle -{{\cos ^{10}\left(x\right)}\over{10}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos \left(x\right)}
\end{array}\]
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