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

Correct Answer is : −(sec⁡x−tan⁡x)323−(sec⁡x−tan⁡x)727+c- {( x - x)³/2}}{3} - {( x - x)^{7/2}}{7} + c−3(secx−tanx)23−7(secx−tanx)27+c

- {( x + x)^{5/2}}{5} - {( x + x)^{7/2}}{7} + c

- {( x - x)^{5/2}}{5} - {( x - x)^{7/2}}{7} + c

- {( x + x)³/2}}{3} - {( x + x)^{7/2}}{7} + c

- {( x - x)³/2}}{3} - {( x - x)^{7/2}}{7} + c

Correct Answer is : −(sec⁡x−tan⁡x)323−(sec⁡x−tan⁡x)727+c- {( x - x)³/2}}{3} - {( x - x)^{7/2}}{7} + c−3(secx−tanx)23−7(secx−tanx)27+c

Test Index

Indefinite Integration

Section: Mathematics

Question : 18 of 89

Marks: +1, -0

\int \frac{\sec^2 x}{(\sec x + \tan x)^{\frac{5}{2}}} dx =

$- \frac{(\sec x + \tan x)^{\frac{5}{2}}}{5} - \frac{(\sec x + \tan x)^{\frac{7}{2}}}{7} + c$
$- \frac{(\sec x - \tan x)^{\frac{5}{2}}}{5} - \frac{(\sec x - \tan x)^{\frac{7}{2}}}{7} + c$
$- \frac{(\sec x + \tan x)^{\frac{3}{2}}}{3} - \frac{(\sec x + \tan x)^{\frac{7}{2}}}{7} + c$
$- \frac{(\sec x - \tan x)^{\frac{3}{2}}}{3} - \frac{(\sec x - \tan x)^{\frac{7}{2}}}{7} + c$

Solution:

Let

I = \int \frac{\sec^2 x}{(\sec x + \tan x)^{\frac{5}{2}}} dx

Put

\sec x + \tan x = t

\therefore \sec x - \tan x = \frac{1}{t}

\therefore \sec x = \frac{1}{2} \left( t + \frac{1}{t} \right)

and

\tan x = \frac{1}{2} \left( t - \frac{1}{t} \right)

Differentiate w.r.t.

x

, we get

\Rightarrow \left( \sec x \tan x + \sec^2 x \right) dx = dt

\Rightarrow dx = \frac{dt}{\sec x \tan x + \sec^2 x}

I = \int \frac{\sec^2 x}{\frac{5}{t^2}} \times \frac{dt}{\sec x \tan x + \sec^2 x}

= \int \frac{\sec x}{t^{\frac{5}{2}}} \times \frac{dt}{(\sec x + \tan x)}

= \int \left( \frac{1}{2} \left( t + \frac{1}{t} \right) \right) \cdot \frac{1}{t^{\frac{5}{2}}} \times \frac{1}{t} dt

= \frac{1}{2} \int \left( t + \frac{1}{t} \right) t^{-\frac{7}{2}} dt

= \frac{1}{2} \left[ \int t^{-\frac{5}{2}} dt + \int t^{-\frac{9}{2}} dt \right]

= \frac{1}{2} \left[ \frac{t^{-\frac{3}{2}}}{-\frac{3}{2}} + \frac{t^{-\frac{7}{2}}}{-\frac{7}{2}} \right] + C

= \frac{1}{2} \left[ -\frac{2}{3} \times \frac{1}{t^{\frac{3}{2}}} - \frac{2}{7} \cdot \frac{1}{\frac{7}{2}} \right] + C

= -\frac{1}{3} (\sec x - \tan x)^{\frac{3}{2}} - \frac{1}{7} (\sec x - \tan x)^{\frac{7}{2}} + C

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