Mathematics

Solve : $$\displaystyle\int sec^{-1} \sqrt{x} \, dx$$


SOLUTION
$$\displaystyle\int \sec^{-1}\sqrt{x}dx$$
Let $$x=\cos^2t$$ here $$t=\cos^{-1}\sqrt{x}$$
$$dx=-2\cos t\sin tdt$$
Put there value
$$-\displaystyle\int 2\sec^{-1}\sqrt{\cos^2x}\cos t\sin t dt$$
$$=-2\displaystyle\int \sec^{-1}\cos x\cos t\sin t dt$$
$$=-2\displaystyle\int x\cos t\sin tdt$$
$$=-2\displaystyle\int \cos^2t\cos t\sin tdt$$
$$=-2\displaystyle\int \cos^3t\sin tdt$$
Let $$\cos t=m$$
$$-\sin tdt=dm$$
$$=-2\displaystyle\int m^3dm$$
$$=-2\left[\dfrac{m^4}{4}\right]+c$$
$$=-\dfrac{\cos^4t}{2}+c=-\dfrac{\cos^4(\cos^{-1}\sqrt{x})}{2}+c$$.
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Subjective Hard Published on 17th 09, 2020
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