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CBSE Class 12 · 2020 · Indefinite IntegralsLong Answer

Find

\int sec^{3} x d x

.

Answer: $\frac{1}{2} sec x tan x + \frac{1}{2} lo g ∣ sec x + tan x ∣ + C$

Solution

Write

sec^{3} x = sec x \cdot sec^{2} x

and integrate by parts with

u = sec x

,

d v = sec^{2} x d x

:

I = sec x tan x - \int sec x tan^{2} x d x = sec x tan x - \int sec x (sec^{2} x - 1) d x .

So

I = sec x tan x - I + \int sec x d x

, giving

2 I = sec x tan x + lo g ∣ sec x + tan x ∣.

\frac{1}{2} sec x tan x + \frac{1}{2} lo g ∣ sec x + tan x ∣ + C

CBSE Class 12 Board · 2020 · Subjective

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