The integral I= displaystyle ∫ [x ex2 (sin x2 + cos ⁡ x2)]dx =f(x)+c, (where, c is the constant of integration). Then, f(x) can be

Question

The integral I= displaystyle ∫ [x ex2 (sin x2 + cos ⁡ x2)]dx =f(x)+c, (where, c is the constant of integration). Then, f(x) can be (A) exsin (x2) (B) ex2sin (x) (C) ex2sin ((x2/2)) (D) (1/2)ex2sin (x2)

Tardigrade Admin · Accepted Answer

Let, x2=t⇒ 2xdx=dt ∴ I=(1/2) displaystyle ∫ et(sin t + cos â¡ t)dt =(1/2)et⋅ sin t+c =(1/2)ex2sin (x2)+c . As, .∫ ex(f(x)+f prime(x)) d x=ex ⋅ f(x)+c

Q. The integral $I = \int [x e^{x^{2}} (s in x^{2} + cos x^{2})] d x$ $= f (x) + c,$ (where, $c$ is the constant of integration). Then, $f (x)$ can be

$e^{x} s in (x^{2})$

$e^{x^{2}} s in (x)$

$e^{x^{2}} s in (\frac{x ^{2}}{2})$

$\frac{1}{2} e^{x^{2}} s in (x^{2})$

Let, $x^{2} = t \Rightarrow 2 x d x = d t$
$∴ I = \frac{1}{2} \int e^{t} (s in t + cos t) d t$
$= \frac{1}{2} e^{t} \cdot s in t + c$
$= \frac{1}{2} e^{x^{2}} s in (x^{2}) + c$
${$ As, $\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} \cdot f (x) + c}$

Q. The integral I=∫[xex2(sinx2+cos⁡x2)]dx =f(x)+c, (where, c is the constant of integration). Then, f(x) can be

exsin(x2)

ex2sin(x)

ex2sin(2x2​)

21​ex2sin(x2)