The value of the integral -02x-12sinx-1-x-12-cosx-1dx is

I=∫_0^2(x 1)^2sin(x 1)(x 1)^2+cos(x 1)dx put x 1=t⇒ dx=dt I=∫_ 1^+1t^2sin tt^2+cos tdt f(t)=t^2sin tt^2+cos t f( t)= f(t) So the function is an odd function. He ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Monotonically Decreasing Functions

The value of ...

Question

The value of the integral $\int_{0}^{2} \frac{(x - 1)^{2} s i n (x - 1)}{(x - 1)^{2} + c o s (x - 1)} d x$ is

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0$
$I = \int_{0}^{2} \frac{(x - 1)^{2} s i n (x - 1)}{(x - 1)^{2} + c o s (x - 1)} d x$
put $x - 1 = t \Rightarrow d x = d t$
$I = \int_{- 1}^{+ 1} \frac{t^{2} s i n t}{t^{2} + c o s t} d t$
$f (t) = \frac{t^{2} s i n t}{t^{2} + c o s t}$
$f (- t) = - f (t)$
So the function is an odd function.
Hence, $I = 0$

Suggest Corrections

The value of the integral ∫20(x−1)2sin(x−1)(x−1)2+cos(x−1)dx is

The correct option is B 0I=∫20(x−1)2sin(x−1)(x−1)2+cos(x−1)dx put x−1=t⇒dx=dt I=∫+1−1t2sintt2+costdt f(t)=t2sintt2+cost f(−t)=−f(t) So the function is an odd function. Hence, I=0

The value of the integral $\int_{0}^{2} \frac{(x - 1)^{2} s i n (x - 1)}{(x - 1)^{2} + c o s (x - 1)} d x$ is