Question

\( \begin{array}{l}x e^{\sin x} \\ x e^{u} \quad u=\sin x \\ x e^{u}+e^{u} \quad \frac{d u}{d x}=\cos x \\ u=x \\ u^{\prime}=1>v=e^{u} \\ =x v^{s}=e^{u} \\ =x \cos x e^{u}+\cos x e^{u} \\ x \cos x e^{\sin x}+\cos x e^{\sin x}\end{array} \)

          \( \begin{array}{l}x e^{\sin x} \\ x e^{u} \quad u=\sin x \\ x e^{u}+e^{u} \quad \frac{d u}{d x}=\cos x \\ u=x \\ u^{\prime}=1>v=e^{u} \\ =x v^{s}=e^{u} \\ =x \cos x e^{u}+\cos x e^{u} \\ x \cos x e^{\sin x}+\cos x e^{\sin x}\end{array} \)

x e^sin x
x e^u u=sin x
x e^u+e^u (d u)/(d x)=cos x
u=x
u^'=1>v=e^u
=x v^s=e^u
=x cos x e^u+cos x e^u
x cos x e^sin x+cos x e^sin x

Added by Christy A.

Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Sanjiva Dayal IIT Kanpur 1st Edition

Chapter 6

Instant Answer

Solved by Expert Linda Winkler

Step 1

The function given is \( x e^{\sin x} \). Show more…

Show all steps

Thanks for your feedback!

\( \begin{array}{l}x e^{\sin x} \\ x e^{u} \quad u=\sin x \\ x e^{u}+e^{u} \quad \frac{d u}{d x}=\cos x \\ u=x \\ u^{\prime}=1>v=e^{u} \\ =x v^{s}=e^{u} \\ =x \cos x e^{u}+\cos x e^{u} \\ x \cos x e^{\sin x}+\cos x e^{\sin x}\end{array} \)