\int{-\frac{30\sin^{-1}{(3x)}}{\sqrt{1-9x^2}}dx}Step By Step Solution:
The given integral is:
\int{-\frac{30\sin^{-1}{(3x)}}{\sqrt{1-9x^2}}dx}Factor the term as:
\int{-\frac{30\sin^{-1}{(3x)}}{\sqrt{1-\textcolor{primary}{(3x)^2}}}dx}Let u=3x by Differentials, it follows that, du=3dx as,
\frac{du}{dx}\implies\frac{d}{dx}(3x)=3Dividing each side by 3 it gives:
\frac{1}{3}du=\frac{{3}}{{3}}dx
\frac{1}{{3}}du=\frac{\cancel{3}}{\cancel{3}}dx
After canceling we get :
\frac{1}{{3}}du=dxNow,
\int{-\frac{30\sin^{-1}{(3x)}}{\sqrt{1-{(3x)^2}}}dx}Substitute
{3x=u} \\ {dx=\frac{1}{3}du}
into the expression:
\int{-\frac{30\sin^{-1}{u}}{\sqrt{1-{u^2}}}{\frac{1}{3}du}}Rewrite the expression using constant property
\frac{30}{3}\int{-\frac{\sin^{-1}{(u)}}{\sqrt{1-{u^2}}}du}Rewrite the expression as:
-\frac{30}{3}\int{ \sin^{-1}{(u)}\cdot (\frac{1}{\sqrt{1-{u^2}}})~du}Recall that
\dfrac{d}{du}(\sin^{-1}(u)) =\frac{1}{\sqrt{1-u^{2}}}Let
z=\sin^{-1}(u)by Differentials, it follows that,
dz=\dfrac{1}{\sqrt{1-u^2}}~du.Now,
-\frac{30}{3}\int{\sin^{-1}{(u)}\cdot {(\frac{1}{\sqrt{1-{u^2}}})~du}}Substitute
{z=\sin^{-1}(u)} \\{dz=\dfrac{1}{\sqrt{1-u^2}}~du}into the expression,through it we get:
-\frac{30}{3}\int{z}~ {dz}Evaluate the integral,
-\frac{30}{3}\cdot\frac{z^2}{2}Add constant as it is indefinite integral, so we will get:
-\frac{30}{3}\cdot\frac{z^2}{2}+C\implies-5{z^2}+CTherefore, The value of
\frac{30}{3}\int{ z}~ dzis:
-\frac{30}{3}\int{ z}~ dz=-5z^{2}+CNow,
-5{z^{2}}+CSubstitute
z=\sin^{-1}(u)into the expression:
-5({\sin^{-1}(u)})^2+CNow, Substitute
u=3x
into the expression:
-5({\sin^{-1}(\textcolor{primary}{3x})})^2+CThe required answer is:
\int{-\frac{30\sin^{-1}{(3x)}}{\sqrt{1-9x^2}}dx}=-5({\sin^{-1}({3x})})^2+CSimilar Post
1 thought on “Evaluate the integral”