Correct Answer is : - csc⁡3xcot⁡x4−38csc⁡xcot⁡x+38log⁡∣tan⁡x2∣+c{³ x x}{4} - 3/8 x x + 3/8 ≤ft| x/2 | + c4csc3xcotx−83cscxcotx+83logtan2x+c

{ x ³ x}{4} - 5/8 x x + 3/8 ≤ft| x/2 | + c

- { x ³ x}{4} - 5/8 x x + 3/8 ≤ft| x/2 | + c

- {³ x x}{4} - 3/8 x x + 3/8 ≤ft| x/2 | + c

- {³ x x}{4} + 3/8 x x - 3/8 ≤ft| x/2 | + c

Correct Answer is : - csc⁡3xcot⁡x4−38csc⁡xcot⁡x+38log⁡∣tan⁡x2∣+c{³ x x}{4} - 3/8 x x + 3/8 ≤ft| x/2 | + c4csc3xcotx−83cscxcotx+83logtan2x+c

Test Index

TS EAMCET 04 May 2019 Shift 2 Solved Paper

Section: Mathematics

Question : 74 of 160

Marks: +1, -0

∫

\csc^5

x dx =

$\frac{\csc x \cot^{3} x}{4} - \frac{5}{8} \csc x \cot x + \frac{3}{8} \log \left| \tan \frac{x}{2} \right| + c$
- $\frac{\csc x \cot^{3} x}{4} - \frac{5}{8} \csc x \cot x + \frac{3}{8} \log \left| \tan \frac{x}{2} \right| + c$
- $\frac{\csc^{3} x \cot x}{4} - \frac{3}{8} \csc x \cot x + \frac{3}{8} \log \left| \tan \frac{x}{2} \right| + c$
- $\frac{\csc^{3} x \cot x}{4} + \frac{3}{8} \csc x \cot x - \frac{3}{8} \log \left| \tan \frac{x}{2} \right| + c$

Solution:

Consider the integral.

I=\int \csc^5 x \, dx

The above integral is solved as

I = \int \csc^{3} x \cdot \csc^{2} x \, dx

= \csc^{3} x (-\cot x) - \int (-\cot x) 3 \csc^{2} x (-\csc x \cot x) \, dx

= -\csc^{3} x \cot x - 3 \int \csc^{3} x \cot^{2} x \, dx

= -\csc^{3} x \cot x - 3 \int \csc^{5} x \, dx + \int \csc^{3} x \, dx

Solve further

I = -\csc^{3} x \cot x - 3I + 3 I_{1}

Here

I_{1} = \int \csc^{3} x \, dx

It is solved as

I_{1} = \int \csc x \cdot \csc^{2} x \, dx

= -\csc x \cot x - \int \csc x \cdot \cot^{2} x \, dx

= -\csc x \cot x - \int \csc^{3} x \, dx + \int \csc x \, dx

= -\csc x \cot x - I_{1} + \log \tan \frac{x}{2}

\text{So}

2 I_{1} = -\csc x \cot x + \log \tan \frac{x}{2}

Therefore

I = -\csc^{3} x \cot x - 3I + 3 \left[ -\frac{1}{2} \csc x \cot x + \log \tan \frac{x}{2} \right]

This implies

I = -\frac{\csc^{3} x \cot x}{4} - \frac{3}{8} \csc x \cot x + \frac{3}{8} \log \left| \tan \frac{x}{2} \right| + c

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