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Worked solution

Use a double-angle identity and an exponential substitution

Find

(i) ∫11+cos⁡2x dx\displaystyle\int\dfrac1{1+\cos2x}\,dx. marks

(ii) ∫e2xex+2 dx\displaystyle\int\dfrac{e^{2x}}{e^x+2}\,dx, using the substitution u=ex+2u=e^x+2. marks

[2013/MI/I/8(i),(iii)]

CJC · Review Problems · Question 2

Answer

(i) 12tan⁡x+C\frac12\tan x+C; (ii) ex+2−2ln⁡∣ex+2∣+Ce^x+2-2\ln|e^x+2|+C.

(i)

Use the source double-angle guidance: 1+cos⁡2x=2cos⁡2x1+\cos2x=2\cos^2x. Therefore the integral is 12∫sec⁡2x dx=12tan⁡x+C\frac12\int\sec^2x\,dx=\frac12\tan x+C.

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(ii)

With u=ex+2u=e^x+2, du=exdx=(u−2)dxdu=e^x dx=(u-2)dx. The integral becomes ∫(u−2)2uduu−2=∫(1−2/u) du=u−2ln⁡∣u∣+C\int\dfrac{(u-2)^2}{u}\dfrac{du}{u-2}=\int(1-2/u)\,du=u-2\ln|u|+C. Substitute back: ex+2−2ln⁡∣ex+2∣+Ce^x+2-2\ln|e^x+2|+C. The absolute-value bars match the key; the argument is actually positive.

How did this part go?

Common mistakes

The added constant 2 in an indefinite antiderivative can be absorbed into C, but is retained to match the printed form.

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