int {e^{x/2}} sin left( frac{x}{2} + frac{pi} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) माना $I = \int {e^{x/2}} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$. सूत्र $\int e^{ax} \sin(bx + c) \, dx = \frac{e^{ax}}{a^2 + b^2}

Vedclass · Accepted Answer

$\sqrt{2} {e^{x/2}} \sin \frac{x}{2} + c$

Vedclass · Answer

(D) माना $I = \int {e^{x/2}} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$. सूत्र $\int e^{ax} \sin(bx + c) \, dx = \frac{e^{ax}}{a^2 + b^2} [a \sin(bx + c) - b \cos(bx + c)] + C$ का उपयोग करते हुए। यहाँ $a = 1/2$ और $b = 1/2$ है। $I = \frac{e^{x/2}}{(1/2)^2 + (1/2)^2} \left[ \frac{1}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) - \frac{1}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right] + C$. $I = \frac{e^{x/2}}{1/4 + 1/4} \cdot \frac{1}{2} \left[ \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) - \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right] + C$. $I = \frac{e^{x/2}}{1/2} \cdot \frac{1}{2} \left[ \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) - \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right] + C$. $I = e^{x/2} \left[ \sin \frac{x}{2} \cos \frac{\pi}{4} + \cos \frac{x}{2} \sin \frac{\pi}{4} - (\cos \frac{x}{2} \cos \frac{\pi}{4} - \sin \frac{x}{2} \sin \frac{\pi}{4}) \right] + C$. चूँकि $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$I = e^{x/2} \left[ \frac{1}{\sqrt{2}} \sin \frac{x}{2} + \frac{1}{\sqrt{2}} \cos \frac{x}{2} - \frac{1}{\sqrt{2}} \cos \frac{x}{2} + \frac{1}{\sqrt{2}} \sin \frac{x}{2} \right] + C$. $I = e^{x/2} \left[ \frac{2}{\sqrt{2}} \sin \frac{x}{2} \right] + C = \sqrt{2} {e^{x/2}} \sin \frac{x}{2} + C$. अतः,सही विकल्प $D$ है।

$\int {e^{x/2}} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx = $

Similar Questions

$\int \frac{13 \cos 2 x-9 \sin 2 x}{3 \cos 2 x-4 \sin 2 x} d x=$

$\int e^{\tan \theta} (\sec \theta - \sin \theta) d\theta$ का मान ज्ञात कीजिए:

किसी भी पूर्णांक $n \geq 2$ के लिए,मान लीजिए $I_n = \int \tan^n x \, dx$ है। यदि $n \geq 2$ के लिए $I_n = \frac{1}{a} \tan^{n-1} x - b I_{n-2}$ है,तो क्रमित युग्म $(a, b)$ बराबर है

$\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module