² θ (θ + θ)² \, dθ

Correct Answer is : (secθ+tanθ)/6 [3+(sec⁡θ+tan⁡θ)2]≤ft[ 3 + (θ + θ)² ][3+(secθ+tanθ)2] + C

(secθ + tanθ)/2 [2+tanθ(secθ+tanθ)]+C

(secθ + tanθ)/3 [2+4tanθ(secθ+tanθ)]+C

(secθ+tanθ)/6 ≤ft[ 3 + (θ + θ)² ] + C

3(secθ+tanθ)/2 [2+tanθ(secθ+tanθ)]+C

Correct Answer is : (secθ+tanθ)/6 [3+(sec⁡θ+tan⁡θ)2]≤ft[ 3 + (θ + θ)² ][3+(secθ+tanθ)2] + C

Test Index

UPSEE Solved Model Paper 7

Question : 147 of 150

Marks: +1, -0

\int \sec^2 \theta (\sec\theta + \tan\theta)^2 \, d\theta

Solution:

I ∫

\sec^2 \theta

(sec θ + tan θ) . (sec θ + tan θ) dθ

Put sec θ + tan θ = y ... (1)

∴ sec θ (tan θ + sec θ) dθ = dy

Now, sec θ - tan θ =

\frac{1}{y}

... (2)

From (1) and (2) , 2 sec θ =

y + \frac{1}{y}

∴ I =

\frac{1}{2} \int y \left( y + \frac{1}{y} \right)

dy =

\frac{1}{2} \left[ \frac{y^3}{3} + y \right]

+ C

\frac{1}{2} \left[ \frac{(\sec\theta + \tan\theta)^3}{3} + (\sec\theta + \tan\theta) \right]

+ C

\frac{\sec\theta + \tan\theta}{6} \left[ (\sec\theta + \tan\theta)^2 + 3 \right]

+ C

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