Search any question & find its solution
Question:
Answered & Verified by Expert
$\int_0^1 \cos ^{-1} x d x=$
Options:
Solution:
2852 Upvotes
Verified Answer
The correct answer is:
$1$
Let
$\begin{aligned}
\mathrm{I} & =\int_0^1\left(\cos ^{-1} x\right)(1) \mathrm{d} x \\
& =\left[\cos ^{-1} x \cdot x-\int\left(\frac{-1}{\sqrt{1-x^2}} \cdot x\right) \mathrm{d} x\right]_0^1 \\
& =\left[x \cdot \cos ^{-1} x+\int \frac{x}{\sqrt{1-x^2}} \mathrm{~d} x\right]_0^1 \\
& =\left[x \cos ^{-1} x-\sqrt{1-x^2}\right]_0^1 \\
& =\left[1 \cdot \cos ^{-1}(1)-\sqrt{1-(1)^2}\right]-\left[0 \cdot \cos ^{-1}(0)-\sqrt{1-0^2}\right] \\
& =0-(-1) \\
& =1
\end{aligned}$
$\begin{aligned}
\mathrm{I} & =\int_0^1\left(\cos ^{-1} x\right)(1) \mathrm{d} x \\
& =\left[\cos ^{-1} x \cdot x-\int\left(\frac{-1}{\sqrt{1-x^2}} \cdot x\right) \mathrm{d} x\right]_0^1 \\
& =\left[x \cdot \cos ^{-1} x+\int \frac{x}{\sqrt{1-x^2}} \mathrm{~d} x\right]_0^1 \\
& =\left[x \cos ^{-1} x-\sqrt{1-x^2}\right]_0^1 \\
& =\left[1 \cdot \cos ^{-1}(1)-\sqrt{1-(1)^2}\right]-\left[0 \cdot \cos ^{-1}(0)-\sqrt{1-0^2}\right] \\
& =0-(-1) \\
& =1
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.