If $P=\int e^{a x} \cdot \sin b x d x$ and $Q=$ $\int e^{a x} \cdot \cos b x \cdot d x$ then $\tan ^{-1} \frac{P}{Q}+\tan ^{-1} \frac{b}{a}=\ldots \ldots$
अगर $P=\int e^{ax} \cdot \sin bxdx$ और $Q=$ $\int e^{ax} \cdot \cos bx \cdot dx$ तो $\tan ^{-1} \frac{P} {Q}+\tan ^{-1} \frac{b}{a}=\ldots \ldots$
$\int \frac{d x}{\left(x^{2}+a^{2}\right)^{\frac{3}{2}}}$ equals:
$\int \frac{d x}{\left(x^{2}+a^{2}\right)^{\frac{3}{2}}}$ =
$\int \frac{x e^{x} \cdot d x}{(x+1)^{2}}=\ldots .$
$\int e^{x} \frac{1+\sin x}{1+\cos x} \cdot d x$ equals
$\int e^{x} \frac{1+\sin x}{1+\cos x} \cdot d x$ =
$\int \frac{x d x}{x^{4}-x^{2}-2}=\ldots \ldots \ldots$
$\int \frac{d x}{5+12 \cos x} \cdot=\frac{1}{\sqrt{c}} \log \frac{\sqrt{a}+\sqrt{b} \tan \frac{x}{2}}{\sqrt{a}-\sqrt{b} \tan \frac{x}{2}}$ then c will be equal to
$\int \frac{dx}{5+12 \cos x} \cdot=\frac{1}{\sqrt{c}} \log \frac{\sqrt{a}+\sqrt{b} \tan \ frac{x}{2}}{\sqrt{a}-\sqrt{b} \tan \frac{x}{2}}$ तो c का मान क्या होगा ?
Value of the integral $\int \frac{\tan ^{-1} x \cdot d x}{\left(1+x^{2}\right)^{\frac{3}{2}}}$ is
$\int \frac{\tan ^{-1} x \cdot d x}{\left(1+x^{2}\right)^{\frac{3}{2}}}$ =
$\int \sin (\log x) \cdot d x=\ldots .$
$\int \tan ^{-1} x \cdot d x$ equals
$\int \tan ^{-1} x \cdot d x$ =
Enter Your Mobile No. To Login/Register
⇐ Go Back to change the mobile no.
Didn't receive OTP? Resend OTP -OR- Voice call Call Again/Resend OTP in 30 Seconds