Q.Evaluate: $\displaystyle\int \dfrac{6x+5}{\sqrt{1-4x-4x^2}}\, dx$ OR If $y = \sin^{-1}\dfrac{1}{2}\left(\sqrt{1+x}+\sqrt{1-x}\right)$ then show that $\dfrac{dy}{dx} = \dfrac{-1}{2\sqrt{1-x^2}}$.

Rationalise the integrand by multiplying by $\sqrt{1-x}/\sqrt{1-x}$, then split and integrate term by term.

Multiply numerator and denominator inside the square root by $(1-x)$:

$\sqrt{\dfrac{1-x}{1+x}}=\sqrt{\dfrac{(1-x)^2}{(1+x)(1-x)}}=\dfrac{1-x}{\sqrt{1-x^2}}=\dfrac{1}{\sqrt{1-x^2}}-\dfrac{x}{\sqrt{1-x^2}}.$

Now integrate each piece: …

Writing $\cos 2x = \cos^2x(1-\tan^2x)$ converts the integrand into $\sec^2x/\sqrt{1-\tan^2x}$, and substituting $t=\tan x$ gives a standard $\int dt/\sqrt{1-t^2}$ integral.

$\dfrac{\sec x}{\sqrt{\cos 2x}} = \dfrac{1}{\cos x\sqrt{\cos 2x}}$.

Use $\cos 2x = \cos^2x - \sin^2x = \cos^2x(1-\tan^2x)$, so $\sqrt{\cos 2x} = \cos x\sqrt{1-\tan^2x}$ (taking $\cos x>0$).

Then $\dfrac{1}{\cos x\cdot\cos x\sqrt{1-\tan^2x}} = \dfrac{\sec^2x}{\sqrt{1-\tan^2x}}$.

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The substitution $u=e^x$ reduces the integral to $\int\frac{du}{u^2+1} = \tan^{-1}u+c$, giving $\tan^{-1}(e^x)+c$.

Let $u = e^x$, so $du = e^x\,dx$.

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