JEE Main 2022 — Mathematics Calculus

Let $\displaystyle I={\int }_{0}^{\frac{\pi }{2}}60\frac{\mathrm{sin}6x}{\mathrm{sin}x}dx$I=∫02π​​60sinxsin6x​dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}60\frac{2\mathrm{sin}3x\mathrm{cos}3x}{\mathrm{sin}x}dx$$⇒I=∫02π​​60sinx2sin3xcos3x​dx

$$\Rightarrow I=60{\int }_{0}^{\frac{\pi }{2}}\frac{2(3\mathrm{sin}x-4{\mathrm{sin}}^{3}x)(4{\mathrm{cos}}^{3}x-3\mathrm{cos}x)}{\mathrm{sin}x}dx$$⇒I=60∫02π​​sinx2(3sinx−4sin3x)(4cos3x−3cosx)​dx

$$\Rightarrow I=120{\int }_{0}^{\frac{\pi }{2}}(3-4{\mathrm{sin}}^{2}x)(4{\mathrm{cos}}^{2}x-3)\mathrm{cos}xdx$$⇒I=120∫02π​​(3−4sin2x)(4cos2x−3)cosxdx

$$\Rightarrow I=120{\int }_{0}^{\frac{\pi }{2}}(3-4{\mathrm{sin}}^{2}x)(1-4{\mathrm{sin}}^{2}x)\mathrm{cos}xdx$$⇒I=120∫02π​​(3−4sin2x)(1−4sin2x)cosxdx

Now let $\mathrm{sin}x=t\Rightarrow \mathrm{cos}xdx=dt$sinx=t⇒cosxdx=dt and limit changes to $0\text{to }1$0to 1,

So, $\displaystyle I=120{\int }_{0}^{1}(3-4{t}^{2})(1-4{t}^{2})dt$I=120∫01​(3−4t2)(1−4t2)dt

$$\Rightarrow I=120{\int }_{0}^{1}3-16{t}^{2}+16{t}^{4}dt$$⇒I=120∫01​3−16t2+16t4dt

$$\Rightarrow I=120{[3t-\frac{16{t}^{3}}{3}+\frac{16{t}^{5}}{5}]}_{0}^{1}$$⇒I=120[3t−316t3​+516t5​]01​

$$\Rightarrow I=120[3-\frac{16}{3}+\frac{16}{5}]=120\times \frac{45-80+48}{15}=8\times 13=104$$⇒I=120[3−316​+516​]=120×1545−80+48​=8×13=104