The value of the definite integral ∫ limits0π((1-x sin 2 x) e cos 2 x+(1+x sin 2 x) e sin 2 x) d x is equal to

Question

The value of the definite integral ∫ limits0π((1-x sin 2 x) e cos 2 x+(1+x sin 2 x) e sin 2 x) d x is equal to (A) π (B) π e (C) π( e -1) (D) π( e +1)

Tardigrade Admin · Accepted Answer

∫(f(x)+x f prime(x)) d x=x f(x)+C So, . I=x(e cos 2 x+e sin 2 x)]0π=π(e+1).

Q. The value of the definite integral $0 \int π ((1 - x sin 2 x) e^{c o s^{2} x} + (1 + x sin 2 x) e^{s i n^{2} x}) d x$ is equal to

$π$

$π e$

$π (e - 1)$

$π (e + 1)$

$\int (f (x) + x f^{'} (x)) d x = x f (x) + C$
So, $I = x (e^{c o s^{2} x} + e^{s i n^{2} x})]_{0}^{π} = π (e + 1)$ .

Q. The value of the definite integral 0∫π​((1−xsin2x)ecos2x+(1+xsin2x)esin2x)dx is equal to

π

πe

π(e−1)

π(e+1)

∫(f(x)+xf′(x))dx=xf(x)+C So, I=x(ecos2x+esin2x)]0π​=π(e+1).

Q. The value of the definite integral $0 \int π ((1 - x sin 2 x) e^{c o s^{2} x} + (1 + x sin 2 x) e^{s i n^{2} x}) d x$ is equal to

$π$

$π e$

$π (e - 1)$

$π (e + 1)$

$\int (f (x) + x f^{'} (x)) d x = x f (x) + C$
So, $I = x (e^{c o s^{2} x} + e^{s i n^{2} x})]_{0}^{π} = π (e + 1)$ .