Search any question & find its solution
Question:
Answered & Verified by Expert
$\int_0^{\pi / 2} \frac{1}{1+\tan ^{2020}(x)} d x=$
Options:
Solution:
1665 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{4}$
$$
I=\int_0^{\pi / 2} \frac{d x}{1+\tan ^{2020} x}=\int_0^{\pi / 2} \frac{\cos ^{2020} x}{\cos ^{2020} x+\sin ^{2020} x} d x
$$
On applying property $\int_0^a f(x) d x=\int_0^a f(a-x) d x$, we get
$$
I=\int_0^{\pi / 2} \frac{\sin ^{2020} x}{\sin ^{2020} x+\cos ^{2020} x} d x
$$
On adding Eqs. (i) and (ii), we get
$$
2 I=\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}
$$
Hence, option (3) is correct.
I=\int_0^{\pi / 2} \frac{d x}{1+\tan ^{2020} x}=\int_0^{\pi / 2} \frac{\cos ^{2020} x}{\cos ^{2020} x+\sin ^{2020} x} d x
$$
On applying property $\int_0^a f(x) d x=\int_0^a f(a-x) d x$, we get
$$
I=\int_0^{\pi / 2} \frac{\sin ^{2020} x}{\sin ^{2020} x+\cos ^{2020} x} d x
$$
On adding Eqs. (i) and (ii), we get
$$
2 I=\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}
$$
Hence, option (3) is correct.
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.