JEE Main 2021 — Mathematics Calculus

$\displaystyle I=\int \frac{\mathrm{sin}\theta \cdot \mathrm{sin}2\theta ({\mathrm{sin}}^{6}\theta +{\mathrm{sin}}^{4}\theta +{\mathrm{sin}}^{2}\theta )\sqrt{2{\mathrm{sin}}^{4}\theta +3{\mathrm{sin}}^{2}\theta +6}}{1-\mathrm{cos}2\theta }d\theta$I=∫1−cos2θsinθ⋅sin2θ(sin6θ+sin4θ+sin2θ)2sin4θ+3sin2θ+6​​dθ$\displaystyle \Rightarrow I=\int \frac{\mathrm{sin}\theta .2\mathrm{sin}\theta \mathrm{cos}\theta \cdot {\mathrm{sin}}^{2}\theta ({\mathrm{sin}}^{4}\theta +{\mathrm{sin}}^{2}\theta +1){(2{\mathrm{sin}}^{4}\theta +3{\mathrm{sin}}^{2}\theta +6)}^{1/2}}{2{\mathrm{sin}}^{2}\theta }d\theta$⇒I=∫2sin2θsinθ.2sinθcosθ⋅sin2θ(sin4θ+sin2θ+1)(2sin4θ+3sin2θ+6)1/2​dθ

$$=\int {\mathrm{sin}}^{2}\theta \cdot \mathrm{cos}\theta ({\mathrm{sin}}^{4}\theta +{\mathrm{sin}}^{2}\theta +1){(2{\mathrm{sin}}^{4}\theta +3{\mathrm{sin}}^{2}\theta +6)}^{1/2}d\theta$$=∫sin2θ⋅cosθ(sin4θ+sin2θ+1)(2sin4θ+3sin2θ+6)1/2dθ

Let $\mathrm{sin}\theta =t\Rightarrow \mathrm{cos}\theta d\theta =dt$sinθ=t⇒cosθdθ=dt

$$\therefore I=\int {t}^{2}({t}^{4}+{t}^{2}+1){(2{t}^{4}+3{t}^{2}+6)}^{1/2}dt$$∴I=∫t2(t4+t2+1)(2t4+3t2+6)1/2dt

$$=\int ({t}^{5}+{t}^{3}+t)t{(2{t}^{4}+3{t}^{2}+6)}^{1/2}dt$$=∫(t5+t3+t)t(2t4+3t2+6)1/2dt

$$=\int ({t}^{5}+{t}^{3}+t){({t}^{2})}^{1/2}{(2{t}^{4}+3{t}^{2}+6)}^{1/2}dt$$=∫(t5+t3+t)(t2)1/2(2t4+3t2+6)1/2dt

$$=\int ({t}^{5}+{t}^{3}+t){(2{t}^{6}+3{t}^{4}+6{t}^{2})}^{1/2}dt$$=∫(t5+t3+t)(2t6+3t4+6t2)1/2dt

Let $2{t}^{6}+3{t}^{4}+6{t}^{2}={u}^{2}$2t6+3t4+6t2=u2

$$\Rightarrow 12({t}^{5}+{t}^{3}+t)dt=2udu$$⇒12(t5+t3+t)dt=2udu

$$\therefore I=\int {({u}^{2})}^{1/2}\cdot \frac{2udu}{12}$$∴I=∫(u2)1/2⋅122udu​

$$=\int \frac{{u}^{2}}{6}du=\frac{{u}^{3}}{18}+C$$=∫6u2​du=18u3​+C

$$=\frac{{(2{t}^{6}+3{t}^{4}+6{t}^{2})}^{3/2}}{18}+C$$=18(2t6+3t4+6t2)3/2​+C

when $t=\mathrm{sin}\theta$t=sinθ

and ${t}^{2}=1-{\mathrm{cos}}^{2}\theta$t2=1−cos2θ will give

$$=\frac{1}{18}{[11-18{\mathrm{cos}}^{2}\theta +9{\mathrm{cos}}^{4}\theta -2{\mathrm{cos}}^{6}\theta ]}^{\frac{3}{2}}+c$$=181​[11−18cos2θ+9cos4θ−2cos6θ]23​+c