Skip to content
Exercise 11.2 · Q4

Q.Integrate the following functions with respect to xx:

(i) sec⁡2x5\sec^{2}\dfrac{x}{5}
(ii) cosec(5x+3)cot⁡(5x+3)\text{cosec}(5x+3)\cot(5x+3)
(iii) 30sec⁡(2−15x)tan⁡(2−15x)30\sec(2-15x)\tan(2-15x)
Puducherry TnboardTextbookSubjectiveImportance★★★★★
6% · 8/129 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Each part matches a standard derivative pattern (sec⁡2\sec^2, coseccot⁡\text{cosec}\cot, sec⁡tan⁡\sec\tan) with a linear argument; integrate and divide by aa, keeping any leading numeric coefficient (part iii) as an overall multiplier.

Step 1. Part (i). Argument x5\dfrac{x}{5} has a=15a=\dfrac15; ∫sec⁡2x dx=tan⁡x+c\int\sec^2x\,dx=\tan x+c, so

∫sec⁡2x5 dx=11/5tan⁡x5+c=5tan⁡x5+c.\int \sec^2\frac{x}{5}\,dx=\frac{1}{1/5}\tan\frac{x}{5}+c=5\tan\frac{x}{5}+c.

Step 2. Part (ii). a=5a=5; ∫cosec xcot⁡x dx=−cosec x+c\int \text{cosec}\,x\cot x\,dx=-\text{cosec}\,x+c, so

∫cosec(5x+3)cot⁡(5x+3) dx=15(−cosec(5x+3))+c=−15cosec(5x+3)+c.\int \text{cosec}(5x+3)\cot(5x+3)\,dx=\frac15\big(-\text{cosec}(5x+3)\big)+c=-\frac15\text{cosec}(5x+3)+c.

Step 3. Part (iii). a=−15a=-15; ∫sec⁡xtan⁡x dx=sec⁡x+c\int\sec x\tan x\,dx=\sec x+c, so first integrate sec⁡(2−15x)tan⁡(2−15x)\sec(2-15x)\tan(2-15x):

∫sec⁡(2−15x)tan⁡(2−15x) dx=1−15sec⁡(2−15x)+c=−115sec⁡(2−15x)+c.\int \sec(2-15x)\tan(2-15x)\,dx = \frac{1}{-15}\sec(2-15x)+c=-\frac{1}{15}\sec(2-15x)+c.

Multiplying by the leading constant 3030:

∫30sec⁡(2−15x)tan⁡(2−15x) dx=30×(−115)sec⁡(2−15x)+c=−2sec⁡(2−15x)+c.\int 30\sec(2-15x)\tan(2-15x)\,dx = 30\times\left(-\frac{1}{15}\right)\sec(2-15x)+c=-2\sec(2-15x)+c. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.