MCQ
The value of $\int_0^{2 x} \cos ^{99} x d x$ is
  • A
    1
  • B
    -1
  • C
    99
  • $0$

Answer

Correct option: D.
$0$
(D)
$\int_0^{2 \pi} \cos ^{99} x d x=2 \int_0^\pi \cos ^{99} x d x$
$\ldots\left[\because \int_0^{2 a } f (x) d x=2 \int_0^{ a } f (x) d x\right.$, if $\left.f (2 a -x)= f (x)\right]$
Let $I _1=\int_0^\pi \cos ^{99} x d x$
$\Rightarrow I _1=-\int_0^\pi \cos ^{99} x d x \ldots .\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\Rightarrow I _1=- I _1 \Rightarrow 2 I _1=0 \Rightarrow I _1=0$
$\therefore \quad \int_0^{2 \pi} \cos ^{99} x d x=2(0)=0$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free