\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)=3

Dividing 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=dx

Now,

\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}+C

Therefore, The value of

\frac{30}{3}\int{ z}~ dz

is:

-\frac{30}{3}\int{ z}~ dz=-5z^{2}+C

Now,

-5{z^{2}}+C

Substitute

z=\sin^{-1}(u)

into the expression:

-5({\sin^{-1}(u)})^2+C

Now, Substitute

u=3x

into the expression:

-5({\sin^{-1}(\textcolor{primary}{3x})})^2+C

**The required answer is:**

\int{-\frac{30\sin^{-1}{(3x)}}{\sqrt{1-9x^2}}dx}=-5({\sin^{-1}({3x})})^2+C

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