\int \frac{ \sec ^2 x}{( \sec x+\tan x)^{5 \/ 2}} \dx= [AP EAPCET 23 May 2025 S1]

Correct Answer is : −( secx−tanx)3∕23−( secx−tanx)7∕27+c

-\frac{( \sec x+\tan x)^{5 \/ 2}}{5}-\frac{( \sec x+\tan x)^{7 \/ 2}}{7}+c

-\frac{( \sec x-\tan x)^{5 \/ 2}}{5}-\frac{( \sec x-\tan x)^{7 \/ 2}}{7}+c

-\frac{( \sec x+\tan x)^{3 \/ 2}}{3}-\frac{( \sec x+\tan x)^{7 \/ 2}}{7}+c

-\frac{( \sec x-\tan x)^{3 \/ 2}}{3}-\frac{( \sec x-\tan x)^{7 \/ 2}}{7}+c

AP EAPCET 23 May 2025 Shift 1 Paper

Section: Mathematics

Question : 71 of 160

Marks: +1, -0

∫‌

sec2x

( secx+tan‌x)5∕2

‌dx=

[AP EAPCET 23 May 2025 S1]

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