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Worked solution

Trigonometric identities and reverse chain rules

Find the following integrals:

(a) ∫sec⁡2(5x)tan⁡3(5x) dx\displaystyle\int \sec^2(5x)\tan^3(5x)\,dx

(b) ∫sin⁡y1+2cos⁡y dy\displaystyle\int \dfrac{\sin y}{1+2\cos y}\,dy

(c) ∫excosec⁡2(ex) dx\displaystyle\int e^x\operatorname{cosec}^2(e^x)\,dx

(d) ∫tan⁡2(2t) dt\displaystyle\int \tan^2(2t)\,dt

(e) ∫11−cos⁡2x dx\displaystyle\int \dfrac1{1-\cos2x}\,dx.

CJC · Discussion · Question 6

Answer

(a) 120tan⁡4(5x)+C\dfrac1{20}\tan^4(5x)+C.

(b) −12ln⁡∣1+2cos⁡y∣+C-\dfrac12\ln|1+2\cos y|+C.

(c) −cot⁡(ex)+C-\cot(e^x)+C.

(d) 12tan⁡(2t)−t+C\dfrac12\tan(2t)-t+C.

(e) −12cot⁡x+C-\dfrac12\cot x+C.

(a)

Insert the inner derivative: 15∫[5sec⁡2(5x)]tan⁡3(5x) dx=15⋅tan⁡4(5x)4\dfrac15\int[5\sec^2(5x)]\tan^3(5x)\,dx=\dfrac15\cdot\dfrac{\tan^4(5x)}4.

The integral is 120tan⁡4(5x)+C\dfrac1{20}\tan^4(5x)+C.

How did this part go?

(b)

Multiply and divide by −2 to obtain −12∫[−2sin⁡y]/[1+2cos⁡y] dy-\dfrac12\int[-2\sin y]/[1+2\cos y]\,dy. The numerator is the denominator derivative.

The integral is −12ln⁡∣1+2cos⁡y∣+C-\dfrac12\ln|1+2\cos y|+C.

How did this part go?

(c)

The inner derivative of eˣ is eˣ and the derivative of −cot u is cosec²u. Apply the reverse chain rule.

The integral is −cot⁡(ex)+C-\cot(e^x)+C.

How did this part go?

(d)

Use tan⁡2(2t)=sec⁡2(2t)−1\tan^2(2t)=\sec^2(2t)-1 and account for the inner derivative 2.

The integral is 12tan⁡(2t)−t+C\dfrac12\tan(2t)-t+C.

How did this part go?

(e)

Use 1−cos⁡2x=2sin⁡2x1-\cos2x=2\sin^2x. The integral is 12∫cosec⁡2x dx\dfrac12\int\operatorname{cosec}^2x\,dx.

The integral is −12cot⁡x+C-\dfrac12\cot x+C.

How did this part go?

Common mistakes

The factors 5, 2 and −2 from inner derivatives must be accounted for.

Read the hint