JEE Main 2019 — Mathematics Calculus

$$I={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{3}x}{\mathrm{sin}x + \mathrm{cos}x}dx...(i)$$I=∫02π​​sinx+cosxsin3x​dx...(i)

Using property of definite integration $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx,$∫ab​f(x)dx=∫ab​f(a+b−x)dx, we get

$$I={\int }_{0}^{\frac{\pi }{2}}\frac{{sin}^{3}(\frac{\pi }{2}-x)}{\mathrm{sin}(\frac{\pi }{2}-x)+\mathrm{cos}(\frac{\pi }{2}-x)} dx$$I=∫02π​​sin(2π​−x)+cos(2π​−x)sin3(2π​−x)​dx

Using \mathrm{sin}(\frac{\pi }{2}-x)=\mathrm{cos}x&\mathrm{cos}(\frac{\pi }{2}-x)=\mathrm{sin}x,

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{cos}}^{3}(x)}{\mathrm{cos}(x)+\mathrm{sin}(x)} dx...(ii)$$⇒I=∫02π​​cos(x)+sin(x)cos3(x)​dx...(ii)

On adding the equation (i)&(ii)

$$\Rightarrow 2I={\int }_{0}^{\frac{\pi }{2}}\frac{{sin}^{3}x+ {cos}^{3}x}{\mathrm{sin}x+\mathrm{cos}x}dx$$⇒2I=∫02π​​sinx+cosxsin3x+cos3x​dx

$$\Rightarrow 2I={\int }_{0}^{\frac{\pi }{2}}\frac{(\mathrm{sin}x+\mathrm{cos}x)({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x-\mathrm{sin}x\mathrm{cos}x)}{\mathrm{sin}x+\mathrm{cos}x}dx$$⇒2I=∫02π​​sinx+cosx(sinx+cosx)(sin2x+cos2x−sinxcosx)​dx

$$\Rightarrow 2I={\int }_{0}^{\frac{\pi }{2}}({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x-\mathrm{sin}x\mathrm{cos}x)dx$$⇒2I=∫02π​​(sin2x+cos2x−sinxcosx)dx

Now, using {\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1&2\mathrm{sin}x\mathrm{cos}x=\mathrm{sin}2x, we get

$$\Rightarrow 2I={\int }_{0}^{\frac{\pi }{2}}(1-\frac{1}{2}\mathrm{sin}2x)dx$$⇒2I=∫02π​​(1−21​sin2x)dx

$$\Rightarrow 2I= {[x+\frac{\mathrm{cos}2x}{4}]}_{0}^{\frac{\pi }{2}}$$⇒2I=[x+4cos2x​]02π​​

$$\Rightarrow 2I=(\frac{\pi }{2}+\frac{\mathrm{cos}2(\frac{\pi }{2})}{4})-(0+\frac{\mathrm{cos}2(0)}{4})$$⇒2I=(2π​+4cos2(2π​)​)−(0+4cos2(0)​)

$$\Rightarrow 2I=(\frac{\pi }{2}-\frac{1}{4})-(\frac{1}{4})$$⇒2I=(2π​−41​)−(41​)

$$\Rightarrow 2I=\frac{\pi }{2}-\frac{1}{2}$$⇒2I=2π​−21​

$$\Rightarrow I=\frac{\pi }{4}-\frac{1}{4}=\frac{\pi -1}{4}.$$⇒I=4π​−41​=4π−1​.