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 \left(t\right)\cdot \sin ^8\left(t\right) \,\dd t=# #{{\sin ^9\left(t\right)}\over{9}} + C#
We apply the substitution method with #g(t)=t^8# and #h(t)=\sin \left(t\right)#, because in that case #g(h(t)) \cdot h'(t)=\cos \left(t\right)\cdot \sin ^8\left(t\right)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos \left(t\right)\cdot \sin ^8\left(t\right) \,\dd t&=& \displaystyle \int \sin ^8\left(t\right) \cdot \cos \left(t\right) \, \dd t \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(t)) \cdot h'(t) \, \dd t \text{ with } h'(t)=\cos \left(t\right)} \\ &=& \displaystyle \int \left(\sin ^8\left(t\right) \right) \, \dd(\sin \left(t\right)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(t)=\dd (h(t))} \\ &=& \displaystyle \int u^8 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\sin \left(t\right)=u} \\ &=& \displaystyle {{u^9}\over{9}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\sin ^9\left(t\right)}\over{9}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\sin \left(t\right)}
\end{array}\]
We apply the substitution method with #g(t)=t^8# and #h(t)=\sin \left(t\right)#, because in that case #g(h(t)) \cdot h'(t)=\cos \left(t\right)\cdot \sin ^8\left(t\right)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \cos \left(t\right)\cdot \sin ^8\left(t\right) \,\dd t&=& \displaystyle \int \sin ^8\left(t\right) \cdot \cos \left(t\right) \, \dd t \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(t)) \cdot h'(t) \, \dd t \text{ with } h'(t)=\cos \left(t\right)} \\ &=& \displaystyle \int \left(\sin ^8\left(t\right) \right) \, \dd(\sin \left(t\right)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(t)=\dd (h(t))} \\ &=& \displaystyle \int u^8 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\sin \left(t\right)=u} \\ &=& \displaystyle {{u^9}\over{9}} +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\sin ^9\left(t\right)}\over{9}} +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\sin \left(t\right)}
\end{array}\]
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