Show definite integral equals specific value (trigonometric substitution or identity)

8 marks Standard +0.8

5

Prove the identity $\tan 2 \theta - \tan \theta \equiv \tan \theta \sec 2 \theta$.
Hence show that $\int _ { 0 } ^ { \frac { 1 } { 6 } \pi } \tan \theta \sec 2 \theta \mathrm {~d} \theta = \frac { 1 } { 2 } \ln \frac { 3 } { 2 }$.

23 marks Hard +2.3

(a) Given that f is a function such that the integrals exist,
1. use the substitution $u = a - x$ to show that
$$\int _ { 0 } ^ { a } \mathrm { f } ( x ) \mathrm { d } x = \int _ { 0 } ^ { a } \mathrm { f } ( a - x ) \mathrm { d } x$$
Hence use symmetry of $\mathrm { f } ( \sin x )$ on the interval $[ 0 , \pi ]$ to show that $$\int _ { 0 } ^ { \pi } x \mathrm { f } ( \sin x ) \mathrm { d } x = \pi \int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { f } ( \sin x ) \mathrm { d } x$$ (b) Use the result of (a)(i) to show that $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { \sin ^ { n } x } { \sin ^ { n } x + \cos ^ { n } x } \mathrm {~d} x$$ is independent of $n$, and find the value of this integral.
(c) (i) Prove that $$\frac { \cos x } { 1 + \cos x } \equiv 1 - \frac { 1 } { 2 } \sec ^ { 2 } \left( \frac { x } { 2 } \right)$$
Hence use the results from (a) to find $$\int _ { 0 } ^ { \pi } \frac { x \sin x } { 1 + \sin x } \mathrm {~d} x$$ (d) Find $$\int _ { 0 } ^ { \pi } \frac { x \sin ^ { 4 } x } { \sin ^ { 4 } x + \cos ^ { 4 } x } \mathrm {~d} x$$

7 marks Standard +0.8

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { \sin 2 \theta } { 1 + \cos \theta } d \theta = 2 - 2 \ln 2$$

7 marks Standard +0.8

Show that $$\int_0^{\pi/2} \frac{\sin 2\theta}{1 + \cos \theta} \, d\theta = 2 - 2\ln 2$$ [7]

8 marks Challenging +1.2

In this question you must use detailed reasoning.

Show that $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1+\sin 2x}{-\cos 2x} dx = \ln(\sqrt{a} + b)$, where $a$ and $b$ are integers to be determined. [6]
Show that $\int_{\frac{\pi}{2}}^{\frac{2\pi}{3}} \frac{1+\sin 2x}{-\cos 2x} dx$ is undefined, explaining your reasoning clearly. [2]