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Integral of sec²θ(secθ+tanθ)² | JEE

JEE Maths question with a full step-by-step solution.

Question
∫sec⁡2θ (sec⁡θ+tan⁡θ)2 dθ\displaystyle\int \sec^{2}\theta\,(\sec\theta+\tan\theta)^{2}\,d\theta equals
Asec⁡θ+tan⁡θ2[2+tan⁡θ(sec⁡θ+tan⁡θ)]+C\dfrac{\sec\theta+\tan\theta}{2}\big[2+\tan\theta(\sec\theta+\tan\theta)\big]+C
Bsec⁡θ+tan⁡θ3[2+4tan⁡θ(sec⁡θ+tan⁡θ)]+C\dfrac{\sec\theta+\tan\theta}{3}\big[2+4\tan\theta(\sec\theta+\tan\theta)\big]+C
Csec⁡θ+tan⁡θ3[2+tan⁡θ(sec⁡θ+tan⁡θ)]+C\dfrac{\sec\theta+\tan\theta}{3}\big[2+\tan\theta(\sec\theta+\tan\theta)\big]+Ccorrect
D3(sec⁡θ+tan⁡θ)2[2+tan⁡θ(sec⁡θ+tan⁡θ)]+C\dfrac{3(\sec\theta+\tan\theta)}{2}\big[2+\tan\theta(\sec\theta+\tan\theta)\big]+C
Solution
Step 1: Let y=sec⁡θ+tan⁡θy=\sec\theta+\tan\theta. Differentiating,
dy=(sec⁡θtan⁡θ+sec⁡2θ) dθ=sec⁡θ(tan⁡θ+sec⁡θ) dθ=sec⁡θ⋅y dθdy=(\sec\theta\tan\theta+\sec^{2}\theta)\,d\theta=\sec\theta(\tan\theta+\sec\theta)\,d\theta=\sec\theta\cdot y\,d\theta
 Step 2: Use the conjugate identity (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=sec⁡2θ−tan⁡2θ=1(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^{2}\theta-\tan^{2}\theta=1, so sec⁡θ−tan⁡θ=1y\sec\theta-\tan\theta=\dfrac{1}{y}. Adding to yy and halving,
sec⁡θ=12 ⁣(y+1y),tan⁡θ=12 ⁣(y−1y)\sec\theta=\dfrac{1}{2}\!\left(y+\dfrac{1}{y}\right),\qquad \tan\theta=\dfrac{1}{2}\!\left(y-\dfrac{1}{y}\right)
 Step 3: Rewrite the integral by grouping one sec⁡θ y\sec\theta\,y as dydy:
I=∫sec⁡2θ y2 dθ=∫(sec⁡θ y) (sec⁡θ y dθ)=∫sec⁡θ y dy=∫12 ⁣(y+1y)y dy=12∫(y2+1) dyI=\int \sec^{2}\theta\,y^{2}\,d\theta=\int (\sec\theta\,y)\,(\sec\theta\,y\,d\theta)=\int \sec\theta\,y\,dy=\int \dfrac{1}{2}\!\left(y+\dfrac{1}{y}\right)y\,dy=\dfrac{1}{2}\int (y^{2}+1)\,dy
 Step 4: Integrate:
=12(y33+y)+C=y6 (y2+3)=\dfrac{1}{2}\left(\dfrac{y^{3}}{3}+y\right)+C=\dfrac{y}{6}\,(y^{2}+3)
 Step 5: Convert to the option form. With ytan⁡θ=y⋅12 ⁣(y−1y)=y2−12y\tan\theta=y\cdot\dfrac{1}{2}\!\left(y-\dfrac{1}{y}\right)=\dfrac{y^{2}-1}{2},
2+tan⁡θ(sec⁡θ+tan⁡θ)=2+ytan⁡θ=2+y2−12=y2+322+\tan\theta(\sec\theta+\tan\theta)=2+y\tan\theta=2+\dfrac{y^{2}-1}{2}=\dfrac{y^{2}+3}{2}
∴y6(y2+3)=y3⋅y2+32=sec⁡θ+tan⁡θ3[2+tan⁡θ(sec⁡θ+tan⁡θ)]+C\therefore \dfrac{y}{6}(y^{2}+3)=\dfrac{y}{3}\cdot\dfrac{y^{2}+3}{2}=\dfrac{\sec\theta+\tan\theta}{3}\big[2+\tan\theta(\sec\theta+\tan\theta)\big]+C
 Correct answer: (3)
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