Question

solve integral 1/2cosx+sin2x dx​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\displaystyle\int\dfrac{1}{2\,cos(x)+sin(2x)}\,dx}$

$\sf{=\displaystyle\int\dfrac{1}{2\,cos(x)+2\,sin(x)\,cos(x)}\,dx}$

$\sf{=\displaystyle\int\dfrac{1}{2\,cos(x)(1+sin(x))}\,dx}$

$\sf{=\displaystyle\int\dfrac{cos(x)}{2\,cos^2(x)(1+sin(x))}\,dx}$

$\sf{=\displaystyle\int\dfrac{cos(x)}{2(1-sin^2(x))(1+sin(x))}\,dx}$

$\sf{Let\,\,\,sin(x)=t}\\\sf{\implies\,cos(x)\,dx=dt}$

So,

$\sf{=\dfrac{1}{2}\displaystyle\int\dfrac{dt}{(1-t^2)(1+t)}}$

$\sf{=\dfrac{1}{2}\displaystyle\int\dfrac{dt}{(1-t)(1+t)(1+t)}}$

$\sf{=\dfrac{1}{2}\displaystyle\int\dfrac{dt}{(1-t)(1+t)^2}}$

$\sf{=\dfrac{1}{2}\cdot\dfrac{1}{2}\displaystyle\int\dfrac{2}{(1-t)(1+t)^2}dt}$

$\sf{=\dfrac{1}{4}\displaystyle\int\dfrac{1-t+1+t}{(1-t)(1+t)^2}dt}$

$\sf{=\dfrac{1}{4}\displaystyle\int\dfrac{(1-t)+(1+t)}{(1-t)(1+t)^2}dt}$

$\sf{=\dfrac{1}{4}\displaystyle\int\dfrac{(1-t)}{(1-t)(1+t)^2}dt+\dfrac{1}{4}\displaystyle\int\dfrac{(1+t)}{(1-t)(1+t)^2}dt}$

$\sf{=\dfrac{1}{4}\displaystyle\int\dfrac{1}{(1+t)^2}dt+\dfrac{1}{4}\displaystyle\int\dfrac{1}{(1-t)(1+t)}dt}$

$\sf{=\dfrac{1}{4}\displaystyle\int\dfrac{1}{(1+t)^2}dt+\dfrac{1}{4}\displaystyle\int\dfrac{1}{1-t^2}dt}$

$\sf{=\dfrac{1}{4}\cdot\dfrac{-1}{1+t}+\dfrac{1}{4}\cdot\dfrac{1}{2}\,ln\left|\dfrac{1+t}{1-t}\right|+C}$

$\sf{=-\dfrac{1}{4(1+t)}+\dfrac{1}{8}\,ln\left|\dfrac{1+t}{1-t}\right|+C}$

$\sf{=-\dfrac{1}{4(1+sin(x))}+\dfrac{1}{8}\,ln\left|\dfrac{1+sin(x)}{1-sin(x)}\right|+C}$

$\sf{=-\dfrac{1}{4(1+sin(x))}+\dfrac{1}{8}\,ln\left|\dfrac{(1+sin(x))(1+sin(x))}{(1-sin(x))(1+sin(x))}\right|+C}$

$\sf{=-\dfrac{1}{4(1+sin(x))}+\dfrac{1}{8}\,ln\left|\dfrac{(1+sin(x))^2}{1-sin^2(x)}\right|+C}$

$\sf{=-\dfrac{1}{4(1+sin(x))}+\dfrac{1}{8}\,ln\left|\dfrac{(1+sin(x))^2}{cos^2(x)}\right|+C}$

$\sf{=-\dfrac{1}{4(1+sin(x))}+\dfrac{1}{8}\,ln\left|\dfrac{1+sin(x)}{cos(x)}\right|^2+C}$

$\sf{=-\dfrac{1}{4(1+sin(x))}+\dfrac{1}{8}\cdot2\,ln\left|\dfrac{1+sin(x)}{cos(x)}\right|+C}$

$\sf{=-\dfrac{1}{4(1+sin(x))}+\dfrac{1}{4}\,ln\left|sec(x)+tan(x)\right|+C}$