$\int \frac{d \theta}{\sqrt{1+\sin \theta}}=\ldots \ldots$
Value of the integral$\int \frac{\sin x d x}{\sqrt{\cos ^{2} x+2 \cos x+2}} \text { is }$
$\int \frac{\sin x d x}{\sqrt{\cos ^{2} x+2 \cos x+2}} \text { }$= . . . . . . .
$\int \frac{\sec ^{4} \theta \cdot d \theta}{\tan \theta}=$. . . . . . . .
$\int \frac{\sec ^{4} \theta \cdot d \theta}{\tan \theta}=$. . . . . . . . . . .
$\int \frac{x \cdot d x}{x^{4}+x^{2}+1}=$
Value of the indefinite integral $\int \frac{\left(x^{4}-1\right) \cdot d x}{x^{2} \sqrt{x^{4}+x^{2}+1}}$ is
अनिश्चित समाकल $\int \frac{\left(x^{4}-1\right) \cdot d x}{x^{2} \sqrt{x^{4}+x^{2}+1}}$ का मान क्या होगा ?
$\int \frac{d z}{(1+\sqrt{z}) \sqrt{z-z^{2}}}=$
Value of the integral $\int \frac{x^{2}}{\sqrt{1+x}} \cdot d x$ will be
समाकलन $\int \frac{x^{2}}{\sqrt{1+x}} \cdot d x$ का मान होगा
$\int \sqrt{\frac{1-x}{1+x}} \cdot d x$ equals
$\int \sqrt{\frac{1-x}{1+x}} \cdot d x$=
Value of the integral$\int \frac{e^{2 x}}{\left(1+e^{x}\right)} \cdot d x$ is.
$\int \frac{e^{2 x}}{\left(1+e^{x}\right)} \cdot d x$ = -
Value of the integral$\int \frac{\sin 2 x \cdot d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}\text { is }$
$\int \frac{\sin 2 x \cdot d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}\text {}$= . . . . . .
If $\mathrm{I}=\int \frac{\sqrt{x^{2}-a^{2}}}{x} d x$ then value of $\mathrm{I}$ will be
अगर $\mathrm{I}=\int \frac{\sqrt{x^{2}-a^{2}}}{x} d x$ तो $\mathrm{I}$ का मान क्या होगा ?
$\int \frac{\sec ^{2} x \cdot \tan x}{\sec ^{2} x-1} \cdot d x=\ldots \ldots \ldots$
If $\int \frac{d x}{x \sqrt{1+x^{2}}}=k\log \{f(x)\} \quad$ where $f(x)=\frac{1+\sqrt{x^{2}+1}}{x}$ then $\mathrm{k}$ equals
अगर $\int \frac{dx}{x \sqrt{1+x^{2}}}=k\log \{f(x)\} \quad$ जहां $f(x)=\frac{1+\sqrt{x^{2}+1}}{x}$ तो $\mathrm{k}$ किसके बराबर होगा ?
$\int \frac{d z}{\sqrt{z}+\sqrt{1+z}}=\ldots .$
Value of the integral $\int \frac{1-x^{2}}{\left(1+x^{2}\right)} \cdot \frac{d x}{\sqrt{1+x^{4}}}$ is
$\int \frac{1-x^{2}}{\left(1+x^{2}\right)} \cdot \frac{d x}{\sqrt{1+x^{4}}}$
If $\int \frac{d x}{\left(1+x^{2}\right)^{\frac{3}{2}}}-f(x)+\mathrm{c}$; then $f(x)$ equals
अगर $\int \frac{d x}{\left(1+x^{2}\right)^{\frac{3}{2}}}-f(x)+\mathrm{c}$; फिर $f(x)$ बराबर
If $\mathrm{I}=\int \frac{(t+1)^{2}}{t\left(1+t^{2}\right)} \cdot d t$ then $\mathrm{I}$ equals.
अगर $\mathrm{I}=\int \frac{(t+1)^{2}}{t\left(1+t^{2}\right)} \cdot dt$ तो $\mathrm{I} $ किसके बराबर होगा ?
$\int \frac{d x}{\sqrt{(x-\alpha)(x-\beta)}}=\ldots \ldots .$
$\int \frac{\cos x}{\sin x+\cos x} \cdot d x$
$\int \frac{\sin ^{2} x d x}{(1+\cos x)^{2}}$ equals
$\int \frac{\sin ^{2} x d x}{(1+\cos x)^{2}}$= . . . . . . . . . .
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