यदि int(sqrt{operatorname{cosec} x+1}) d x=k | vedclass

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

Question

(A) माना $I = \int \sqrt{\operatorname{cosec} x + 1} dx = \int \sqrt{\frac{1}{\sin x} + 1} dx = \int \sqrt{\frac{1 + \sin x}{\sin x}} dx$.$\sin x = \f

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) माना $I = \int \sqrt{\operatorname{cosec} x + 1} dx = \int \sqrt{\frac{1}{\sin x} + 1} dx = \int \sqrt{\frac{1 + \sin x}{\sin x}} dx$.$\sin x = \frac{2 	an(x/2)}{1 + 	an^2(x/2)}$ और $1 + \sin x = (\sin(x/2) + \cos(x/2))^2$ का उपयोग करके, हम $u = 	an(x/2)$ और $dx = \frac{2}{1+u^2} du$ प्रतिस्थापन करते हैं।समाकलन $I = \int \sqrt{\frac{(1+u)^2}{2u}} \cdot \frac{2}{1+u^2} du = \sqrt{2} \int \frac{1+u}{\sqrt{u}(1+u^2)} du$ बन जाता है।माना $v = \sqrt{u}$, तो $dv = \frac{1}{2\sqrt{u}} du$, इसलिए $du = 2v dv$.$I = 2\sqrt{2} \int \frac{1+v^2}{1+v^4} dv = \sqrt{2} \int \frac{1+1/v^2}{v^2+1/v^2} dv = \sqrt{2} \int \frac{d(v-1/v)}{(v-1/v)^2 + 2} + \sqrt{2} \int \frac{d(v+1/v)}{(v+1/v)^2 - 2}$.इसका मूल्यांकन करने पर $I = 2 	an^{-1}\left(\frac{\sqrt{2	an(x/2)}}{1-	an(x/2)}ight) + c$ प्राप्त होता है।अतः, $k = 2$ और $f(x) = \frac{\sqrt{2	an(x/2)}}{1-	an(x/2)}$.$x = \pi/6$ के लिए, $	an(x/2) = 	an(\pi/12) = 2 - \sqrt{3}$.$f(\pi/6) = \frac{\sqrt{2(2-\sqrt{3})}}{1-(2-\sqrt{3})} = \frac{\sqrt{4-2\sqrt{3}}}{\sqrt{3}-1} = \frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{3}-1} = 1$.इसलिए, $\frac{1}{k} f(\pi/6) = \frac{1}{2} \cdot 1 = \frac{1}{2}$.

यदि $\int(\sqrt{\operatorname{cosec} x+1}) d x=k \tan ^{-1}(f(x))+c$ है, तो $\frac{1}{k} f\left(\frac{\pi}{6}\right)=$

Similar Questions

यदि $\int \sin (101 x)(\sin x)^{99} d x=\frac{\sin (100 x)(\sin x)^\lambda}{\mu}+c$ है, तो $\frac{\lambda}{\mu}=$

$\int \frac{d x}{\sqrt{\left(5+2 x+x^2\right)^3}}$ का मान ज्ञात कीजिए।

यदि $\int \frac{\sqrt{1-x^4}}{x^7} d x=f(x)\left\{\sqrt{1-x^4}\right\}^n+C$ है,तो $(f(x))^n$ का मान ज्ञात कीजिए।

$\int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ का मान है

यदि $\int \sqrt{\frac{2}{1+\sin x}} dx = 2 \log |A(x) - B(x)| + C$ और $0 \leq x \leq \frac{\pi}{2}$ है,तो $B(\frac{\pi}{4}) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module