Question

$\red{ \bf \: evaluate \: this \: : }\\ \\ \pink{\leadsto \: }\orange{ \displaystyle \int \sf \: \frac{sin \: x + cos \: x}{ \sqrt{cos2x} } \: \: dx}$​

Answer 1

We split the integral as,

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{\sin x\ dx}{\sqrt{cos(2x)}}+\int\dfrac{\cos x\ dx}{\sqrt{cos(2x)}} }$

Since $\small\cos(2x)=2\cos^2x-1=1-2\sin^2x,$

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{\sin x\ dx}{\sqrt{2\cos^2x-1}}+\int\dfrac{\cos x\ dx}{\sqrt{1-2\sin^2x}} }$

Consider,

$\small\text{ \displaystyle\longrightarrow I_1=\int\dfrac{\sin x\ dx}{\sqrt{2\cos^2x-1}} }$

$\small\text{ \displaystyle\longrightarrow I_1=-\int\dfrac{-\sin x\ dx}{\sqrt{\left(\sqrt2\,\cos x\right)^2-1^2}} }$

We have,

$\small\text{ \displaystyle\int\dfrac{dx}{\sqrt{(ax+b)^2-c^2}}=\dfrac{1}{a}\log\left|(ax+b)+\sqrt{(ax+b)^2-c^2}\right|+C }$

Then,

$\small\displaystyle\longrightarrow I_1=-\dfrac{1}{\sqrt2}\,\log\left|\sqrt2\,\cos x+\sqrt{2\cos^2x-1}\right|+C_1$

$\small\displaystyle\longrightarrow I_1=-\dfrac{1}{\sqrt2}\,\log\left|\sqrt2\,\cos x+\sqrt{\cos(2x)}\right|+C_1$

Consider,

$\small\text{ \displaystyle\longrightarrow I_2=\int\dfrac{\cos x\ dx}{\sqrt{1-2\sin^2x}} }$

$\small\text{ \displaystyle\longrightarrow I_2=\int\dfrac{\cos x\ dx}{\sqrt{1^2-\left(\sqrt2\,\sin x\right)^2}} }$

We have,

$\small\text{ \displaystyle\int\dfrac{dx}{\sqrt{c^2-(ax+b)^2}}=\dfrac{1}{a}\sin^{-1}\left(\dfrac{ax+b}{c}\right)+C }$

Then,

$\small\displaystyle\longrightarrow I_2=\dfrac{1}{\sqrt2}\sin^{-1}\left(\sqrt2\,\sin x\right)+C_2$

Hence,

$\small\displaystyle\longrightarrow I=\dfrac{1}{\sqrt2}\sin^{-1}\left(\sqrt2\,\sin x\right)-\dfrac{1}{\sqrt2}\log\left|\sqrt2\,\cos x+\sqrt{\cos(2x)}\right|+C_1+C_2$

Taking $\smallC_1+C_2=C,$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{\sqrt2}\left[\sin^{-1}\left(\sqrt2\,\sin x\right)-\log\left|\sqrt2\,\cos x+\sqrt{\cos(2x)}\right|\right]+C}} }$