Indefinite IntegrationhardFree

∫sec²θ(secθ+tanθ)² via secθ form | 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[1+sec⁡θ(sec⁡θ+tan⁡θ)]+C\dfrac{\sec\theta+\tan\theta}{3}\big[1+\sec\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 t=sec⁡θ+tan⁡θt=\sec\theta+\tan\theta. Then
dt=(sec⁡θtan⁡θ+sec⁡2θ) dθ=sec⁡θ(sec⁡θ+tan⁡θ) dθ=sec⁡θ⋅t dθdt=(\sec\theta\tan\theta+\sec^{2}\theta)\,d\theta=\sec\theta(\sec\theta+\tan\theta)\,d\theta=\sec\theta\cdot t\,d\theta
 Step 2: From (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1, get sec⁡θ−tan⁡θ=1t\sec\theta-\tan\theta=\dfrac{1}{t}, hence sec⁡θ=12 ⁣(t+1t)\sec\theta=\dfrac{1}{2}\!\left(t+\dfrac{1}{t}\right). Step 3: Write the integral grouping one sec⁡θ t\sec\theta\,t as dtdt:
I=∫sec⁡2θ t2 dθ=∫(sec⁡θ t) (sec⁡θ t dθ)=∫sec⁡θ t dt=∫12 ⁣(t+1t)t dt=12∫(t2+1) dtI=\int \sec^{2}\theta\,t^{2}\,d\theta=\int (\sec\theta\,t)\,(\sec\theta\,t\,d\theta)=\int \sec\theta\,t\,dt=\int \dfrac{1}{2}\!\left(t+\dfrac{1}{t}\right)t\,dt=\dfrac{1}{2}\int (t^{2}+1)\,dt
 Step 4: Integrate:
=12 ⁣(t33+t)+C=t6 (t2+3)=\dfrac{1}{2}\!\left(\dfrac{t^{3}}{3}+t\right)+C=\dfrac{t}{6}\,(t^{2}+3)
 Step 5: Convert to the sec⁡θ\sec\theta-option form. Using t2=(sec⁡θ+tan⁡θ)2=2sec⁡2θ+2sec⁡θtan⁡θ−1t^{2}=(\sec\theta+\tan\theta)^{2}=2\sec^{2}\theta+2\sec\theta\tan\theta-1,
1+sec⁡θ(sec⁡θ+tan⁡θ)=1+sec⁡2θ+sec⁡θtan⁡θ=t2+321+\sec\theta(\sec\theta+\tan\theta)=1+\sec^{2}\theta+\sec\theta\tan\theta=\dfrac{t^{2}+3}{2}
∴t6(t2+3)=t3⋅t2+32=sec⁡θ+tan⁡θ3[1+sec⁡θ(sec⁡θ+tan⁡θ)]+C\therefore \dfrac{t}{6}(t^{2}+3)=\dfrac{t}{3}\cdot\dfrac{t^{2}+3}{2}=\dfrac{\sec\theta+\tan\theta}{3}\big[1+\sec\theta(\sec\theta+\tan\theta)\big]+C
 Correct answer: (3)
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