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 \sin ^3\left(y\right) \,\dd y=# #{{\cos ^3\left(y\right)}\over{3}}-\cos \left(y\right) + C#
We apply the substitution method with #g(y)=y^2-1# and #h(y)=\cos \left(y\right)#, because in that case #g(h(y)) \cdot h'(y)=\sin ^3\left(y\right)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \sin ^3\left(y\right) \,\dd y&=& \displaystyle \int \left(\cos ^2\left(y\right)-1\right) \cdot -\sin \left(y\right) \, \dd y \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(y)) \cdot h'(y) \, \dd y \text{ with } h'(y)=-\sin \left(y\right)} \\ &&\phantom{xxx}\blue{\text{used the trigonometric rule }\sin^2(y)=1-\cos^2(y)} \\ &=& \displaystyle \int \left(\cos ^2\left(y\right)-1 \right) \, \dd(\cos \left(y\right)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int u^2-1 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos \left(y\right)=u} \\ &=& \displaystyle {{u^3}\over{3}}-u +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\cos ^3\left(y\right)}\over{3}}-\cos \left(y\right) +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos \left(y\right)}
\end{array}\]
We apply the substitution method with #g(y)=y^2-1# and #h(y)=\cos \left(y\right)#, because in that case #g(h(y)) \cdot h'(y)=\sin ^3\left(y\right)# applies. This goes as follows:
\[\begin{array}{rcl}\displaystyle \int \sin ^3\left(y\right) \,\dd y&=& \displaystyle \int \left(\cos ^2\left(y\right)-1\right) \cdot -\sin \left(y\right) \, \dd y \\&&\phantom{xxx}\blue{\text{step 2: rewritten to the form }\int g(h(y)) \cdot h'(y) \, \dd y \text{ with } h'(y)=-\sin \left(y\right)} \\ &&\phantom{xxx}\blue{\text{used the trigonometric rule }\sin^2(y)=1-\cos^2(y)} \\ &=& \displaystyle \int \left(\cos ^2\left(y\right)-1 \right) \, \dd(\cos \left(y\right)) \\ &&\phantom{xxx}\blue{\text{step 3: rewritten using }h'(y)=\dd (h(y))} \\ &=& \displaystyle \int u^2-1 \, \dd u \\ &&\phantom{xxx}\blue{\text{step 4: substituted }\cos \left(y\right)=u} \\ &=& \displaystyle {{u^3}\over{3}}-u +C \\ &&\phantom{xxx}\blue{\text{step 5: found the antiderivative}} \\ &=& \displaystyle {{\cos ^3\left(y\right)}\over{3}}-\cos \left(y\right) +C \\ &&\phantom{xxx}\blue{\text{stap 6: substituted }u=\cos \left(y\right)}
\end{array}\]
Unlock full access
Teacher access
Request a demo account. We will help you get started with our digital learning environment.
Student access
Is your university not a partner?
Get access to our courses via Pass Your Math independent of your university. See pricing and more.
Or visit omptest.org if jou are taking an OMPT exam.
Or visit omptest.org if jou are taking an OMPT exam.