int sec ^{-1} x , dx = | vedclass

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Question

(B) To evaluate the integral $I = \int \sec^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \sec

Vedclass · Accepted Answer

$x \sec ^{-1} x - \log \left| x + \sqrt{x^2 - 1} \right| + c$

Vedclass · Answer

(B) To evaluate the integral $I = \int \sec^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \sec^{-1} x$ and $dv = dx$. Then $du = \frac{1}{|x| \sqrt{x^2 - 1}} \, dx$ and $v = x$. Applying the formula: $I = x \sec^{-1} x - \int x \cdot \frac{1}{|x| \sqrt{x^2 - 1}} \, dx$. Assuming $x > 1$,we have $|x| = x$,so: $I = x \sec^{-1} x - \int \frac{1}{\sqrt{x^2 - 1}} \, dx$. Using the standard integral formula $\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \log \left| x + \sqrt{x^2 - a^2} \right| + C$,we get: $I = x \sec^{-1} x - \log \left| x + \sqrt{x^2 - 1} \right| + C$.

$\int \sec ^{-1} x \, dx =$

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