¿Cómo encontrar la integral $\int \frac {dx}{\sin^2 x + \tan^2x}$?

Question

He tratado de convertir todo en términos de $\tan \frac{x}{2}$. Pero estoy atrapado.

Answer 1

No $t=\tan(x)$, $dx=dt/(1+t^2)$ y $\sin^2(x)=t^2/(1+t^2)$. ¿Por lo tanto, $$\int \frac {dx}{\sin^2 x + \tan^2x}=\int \frac {dt}{(t^2+ (1+t^2)t^2)}=\int \frac {dt}{t^2(2+t^2)}=\frac{1}{2}\int \frac {dt}{t^2}-\frac{1}{2}\int \frac {dt}{2+t^2}.$ $ se puede tomar desde aquí?

P.D. Por que $t=\tan(x/2)$% y $dx=2dt/(1+t^2)$, $\sin(x)=2t/(1+t^2)$y $\tan(x)=2t/(1-t^2)$. Entonces obtenemos la integral de una función racional más complicada.

Answer 2

$$\dfrac1{\sin^2x+\tan^2x}=\dfrac{\cos^2x}{\sin^2x(1+\cos^2x)}$$

Método de$#1:$ % $ $$=\dfrac{\cos^2x+1-1}{\sin^2x(1+\cos^2x)}=\csc^2x-\dfrac1{(1-\cos^2x)(1+\cos^2x)}$

Ahora $$\dfrac2{(1-\cos^2x)(1+\cos^2x)}=\dfrac1{1-\cos^2x}+\dfrac1{1+\cos^2x}$ $

Finalmente $$\dfrac1{1+\cos^2x}=\dfrac{\sec^2x}{2+\tan^2x}$ $

Método de$#2:$

$$\dfrac{\cos^2x}{\sin^2x(1+\cos^2x)}=\dfrac c{(1-c)(1+c)}=\dfrac A{1-c}+\dfrac B{1+c}$$ using Partial Fraction Decomposition where $c=\cos^2x$

Answer 3

$$\begin{align}\int\dfrac{dx}{\sin^2{(x)}+\tan^2{(x)}}\cdot\dfrac{\sec^2{x}}{\sec^2{x}}&=\int\dfrac{\sec^2{x}\ dx}{\tan^2{x}+\sec^2{x}\tan^2{x}}\ &=\int\dfrac{\sec^2{x}\ dx}{\tan^2{x}+(\tan^2{x}+1)\tan^2{x}}\ &=\int\dfrac{\sec^2{x}\ dx}{2\tan^2{x}+\tan^4{x}}\end {Alinee el} $$

Que $u=\tan{x}$ y $du=\sec^2{x}\ dx$

$$\begin{align} &=\int\dfrac{du}{u^4+2u^2}\ &=\int\left(\dfrac{1}{2u^2}-\dfrac{1}{2(u^2+2)}\right)du\ &=-\dfrac{1}{2}\int\dfrac{du}{u^2+2}+\dfrac{1}{2}\int\dfrac{du}{u^2}\ &=-\dfrac{1}{2}\int\dfrac{du}{2(\frac{u^2}{2}+1)}+\dfrac{1}{2}\int\dfrac{du}{u^2}\ &=-\dfrac{1}{4}\int\dfrac{du}{\frac{u^2}{2}+1}+\dfrac{1}{2}\int\dfrac{du}{u^2}\end {Alinee el} $$

Que $s=\frac{u}{\sqrt{2}}$ y $ds=\frac{du}{\sqrt{2}}$.

\begin{align} &=-\dfrac{1}{2\sqrt{2}}\int\dfrac{ds}{s^2+1}+\dfrac{1}{2}\int\dfrac{du}{u^2}\ &=-\frac{\arctan{s}}{2\sqrt{2}}+\dfrac{1}{2}\int\dfrac{du}{u^2}\ &=-\frac{\arctan{s}}{2\sqrt{2}}-\frac{1}{2u}+C\ &=-\frac{\sqrt{2}u\arctan{\frac{u}{\sqrt{2}}+2}}{4u}+C\ &=-\frac{1}{4}\left(\sqrt{2}\arctan{\frac{\tan{x}}{\sqrt{2}}\tan{x}+2} \right)\cot{x}+C \end {Alinee el}

Que es igual a $$-\frac{1}{4}\left(\sqrt{2}\arctan{\frac{\tan{x}}{\sqrt{2}}+2\cot{x}}\right)+C.$ $