Question

❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​​​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{cos(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\,d\theta\,\,\,\,\,\,\,\,\,\,...(1)}$

We know,

$\displaystyle\boxed{\bf{\blue{\int^{b}_{a}\,f(x)\,dx=\int^{b}_{a}\,f(a+b-x)\,dx}}}$

So,

$\displaystyle\tt{I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{cos\left(\dfrac{\pi}{2}-\theta\right)}}{\left(\sqrt{cos\left(\dfrac{\pi}{2}-\theta\right)}+\sqrt{sin\left(\dfrac{\pi}{2}-\theta\right)}\right)^5}\,d\theta}$

$\displaystyle\tt{\implies\,I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{sin(\theta)}}{\left(\sqrt{sin(\theta)}+\sqrt{cos(\theta)}\right)^5}\,d\theta\,\,\,\,\,\,\,\,\,...(2)}$

Add (1) and (2),

$\displaystyle\tt{2I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{cos(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\,d\theta\,\,+\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{sin(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\,d\theta}$

$\displaystyle\tt{\implies2I=\int^{\frac{\pi}{2}}_{0}\,\left[\dfrac{3\,\sqrt{cos(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}+\dfrac{3\,\sqrt{sin(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\right]\,d\theta}$

$\displaystyle\tt{\implies2I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\,d\theta}$

$\displaystyle\tt{\implies2I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^4}\,d\theta}$

$\displaystyle\tt{\implies2I=3\,\int^{\frac{\pi}{2}}_{0}\,\dfrac{1}{cos^2(\theta)\left(1+\sqrt{tan(\theta)}\right)^4}\,d\theta}$

$\displaystyle\tt{\implies2I=3\,\int^{\frac{\pi}{2}}_{0}\,\dfrac{sec^2(\theta)}{\left(1+\sqrt{tan(\theta)}\right)^4}\,d\theta}$

$\bf{Put\,\,\,tan(\theta)=t^2}$

$\bf{\implies\,sec^2(\theta)\,d\theta=2t\,dt}$

So,

$\bf{When\,\,\,\theta=0,\,\,t=0}\\\bf{When\,\,\,\theta=\dfrac{\pi}{2},\,\,t=\infty}$

$\displaystyle\tt{\implies2I=3\,\int^{\infty}_{0}\,\dfrac{2t}{\left(1+t\right)^4}\,dt}$

$\displaystyle\tt{\implies2I=3\,\int^{\infty}_{0}\,\dfrac{2t+2-2}{\left(1+t\right)^4}\,dt}$

$\displaystyle\tt{\implies2I=3\,\int^{\infty}_{0}\,\dfrac{2(t+1)}{\left(1+t\right)^4}\,dt-6\int^{\infty}_{0}\,\dfrac{1}{\left(1+t\right)^4}\,dt}$

$\displaystyle\tt{\implies2I=6\,\int^{\infty}_{0}\,\dfrac{1}{\left(1+t\right)^3}\,dt-6\int^{\infty}_{0}\,\dfrac{1}{\left(1+t\right)^4}\,dt}$

$\displaystyle\tt{\implies2I=-6\left[\dfrac{1}{2\left(1+t\right)^2}\right]^{\infty}_{0}+6\left[\dfrac{1}{3\left(1+t\right)^3}\right]^{\infty}_{0}}$

$\displaystyle\tt{\implies2I=-3\left[\dfrac{1}{(1+t)^2}\right]^{\infty}_{0}+2\left[\dfrac{1}{(1+t)^3}\right]^{\infty}_{0}}$

$\displaystyle\tt{\implies2I=-3(0-1)+2(0-1)}$

$\displaystyle\tt{\implies2I=3-2}$

$\displaystyle\tt{\implies2I=1}$

$\displaystyle\tt{\implies\,I=\dfrac{1}{2}}$

Answer 2

Answer:

We have,

$\displaystyle\tt{I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{cos(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\,d\theta\,\,\,\,\,\,\,\,\,\,...(1)}$

We know,

$\displaystyle\boxed{\bf{\blue{\int^{b}_{a}\,f(x)\,dx=\int^{b}_{a}\,f(a+b-x)\,dx}}}$

So,

$\displaystyle\tt{I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{cos\left(\dfrac{\pi}{2}-\theta\right)}}{\left(\sqrt{cos\left(\dfrac{\pi}{2}-\theta\right)}+\sqrt{sin\left(\dfrac{\pi}{2}-\theta\right)}\right)^5}\,d\theta}$

$\displaystyle\tt{\implies\,I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{sin(\theta)}}{\left(\sqrt{sin(\theta)}+\sqrt{cos(\theta)}\right)^5}\,d\theta\,\,\,\,\,\,\,\,\,...(2)}$

Add (1) and (2),

$\displaystyle\tt{2I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{cos(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\,d\theta\,\,+\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\,\sqrt{sin(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\,d\theta}$

$\displaystyle\tt{\implies2I=\int^{\frac{\pi}{2}}_{0}\,\left[\dfrac{3\,\sqrt{cos(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}+\dfrac{3\,\sqrt{sin(\theta)}}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\right]\,d\theta}$

$\displaystyle\tt{\implies2I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^5}\,d\theta}$

$\displaystyle\tt{\implies2I=\int^{\frac{\pi}{2}}_{0}\,\dfrac{3}{\left(\sqrt{cos(\theta)}+\sqrt{sin(\theta)}\right)^4}\,d\theta}$

$\displaystyle\tt{\implies2I=3\,\int^{\frac{\pi}{2}}_{0}\,\dfrac{1}{cos^2(\theta)\left(1+\sqrt{tan(\theta)}\right)^4}\,d\theta}$

$\displaystyle\tt{\implies2I=3\,\int^{\frac{\pi}{2}}_{0}\,\dfrac{sec^2(\theta)}{\left(1+\sqrt{tan(\theta)}\right)^4}\,d\theta}$

$\bf{Put\,\,\,tan(\theta)=t^2}$

$\bf{\implies\,sec^2(\theta)\,d\theta=2t\,dt}$

So,

$\bf{When\,\,\,\theta=0,\,\,t=0}\\\bf{When\,\,\,\theta=\dfrac{\pi}{2},\,\,t=\infty}$

$\displaystyle\tt{\implies2I=3\,\int^{\infty}_{0}\,\dfrac{2t}{\left(1+t\right)^4}\,dt}$

$\displaystyle\tt{\implies2I=3\,\int^{\infty}_{0}\,\dfrac{2t+2-2}{\left(1+t\right)^4}\,dt}$

$\displaystyle\tt{\implies2I=3\,\int^{\infty}_{0}\,\dfrac{2(t+1)}{\left(1+t\right)^4}\,dt-6\int^{\infty}_{0}\,\dfrac{1}{\left(1+t\right)^4}\,dt}$

$\displaystyle\tt{\implies2I=6\,\int^{\infty}_{0}\,\dfrac{1}{\left(1+t\right)^3}\,dt-6\int^{\infty}_{0}\,\dfrac{1}{\left(1+t\right)^4}\,dt}$

$\displaystyle\tt{\implies2I=-6\left[\dfrac{1}{2\left(1+t\right)^2}\right]^{\infty}_{0}+6\left[\dfrac{1}{3\left(1+t\right)^3}\right]^{\infty}_{0}}$

$\displaystyle\tt{\implies2I=-3\left[\dfrac{1}{(1+t)^2}\right]^{\infty}_{0}+2\left[\dfrac{1}{(1+t)^3}\right]^{\infty}_{0}}$

$\displaystyle\tt{\implies2I=-3(0-1)+2(0-1)}$

$\displaystyle\tt{\implies2I=3-2}$

$\displaystyle\tt{\implies2I=1}$

$\displaystyle\tt{\implies\,I=\dfrac{1}{2}}$