Question

Prove the following



​

Answer 1

Question :-

Using the Principle of Mathematical induction, to prove that

$\rm \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} nx}{sinx} = 1 + \frac{1}{3} + \frac{1}{5} + - - + \frac{1}{2n - 1}$

$\large\underline{\sf{Solution-}}$

Given statement is

$\rm \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} nx}{sinx} = 1 + \frac{1}{3} + \frac{1}{5} + - - + \frac{1}{2n - 1}$

Let assume that,

$\rm \: P(n) \: : \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} nx}{sinx} = 1 + \frac{1}{3} + \frac{1}{5} + - - + \frac{1}{2n - 1}$

Step :- 1 For n = 1

$\rm \: P(1) \: : \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} x}{sinx} = 1$

$\rm \: \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm sinx = 1$

$\rm \: \: \bigg( - cosx\bigg) _{0}^{ \dfrac{\pi}{2} } = 1$

$\rm \: - cos\dfrac{\pi}{2} + cos0 = 1$

$\rm \: - 0 + 1 = 1$

$\rm \: 1 = 1$

$\rm\implies \:P(n) \: is \: true \: for \: n \: = \: 1$

Step :- 2 Let assume that P(n) is true for n = k, where k is a natural number.

$\rm \: P(k) \: : \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} kx}{sinx} = 1 + \frac{1}{3} + \frac{1}{5} + - - + \frac{1}{2k - 1}$

Step :- 3 Now, we have to show that P(n) is true for n = k + 1

$\rm \: P(k + 1) \: : \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} (k + 1)x}{sinx} = 1 + \frac{1}{3} + \frac{1}{5} + - - + \frac{1}{2k + 1}$

Now, Consider LHS

$\rm \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} (k + 1)x}{sinx}$

$\rm \: = \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} (k + 1)x - {sin}^{2} kx + {sin}^{2} kx}{sinx} \: dx$

$\rm \: = \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} (k + 1)x - {sin}^{2} kx }{sinx}dx + \displaystyle\int_{0}^{\dfrac{\pi}{2}}\rm \frac{ {sin}^{2} kx}{sinx}dx$

We know,

$\boxed{\tt{ \: {sin}^{2}x - {sin}^{2}y = sin(x + y) \: sin(x - y) \: }}$

So, using this identity, we get

$\rm \: = \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{sin(2k + 1)x \: sinx}{sinx}dx + \displaystyle\int_{0}^{\dfrac{\pi}{2}}\rm \frac{ {sin}^{2} kx}{sinx}dx$

$\rm \: = \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm sin(2k + 1)xdx + \displaystyle\int_{0}^{\dfrac{\pi}{2}}\rm \frac{ {sin}^{2} kx}{sinx}dx$

$\rm \: = \: - \bigg(\dfrac{cos(2k + 1)x}{2k + 1} \bigg) _{0}^{ \dfrac{\pi}{2} } + \displaystyle\int_{0}^{\dfrac{\pi}{2}}\rm \frac{ {sin}^{2} kx}{sinx}dx$

$\rm \: = \: - \bigg(\dfrac{cos(2k + 1)\dfrac{\pi}{2}}{2k + 1} - \frac{cos0}{2k + 1} \bigg) + \displaystyle\int_{0}^{\dfrac{\pi}{2}}\rm \frac{ {sin}^{2} kx}{sinx}dx$

$\rm \: = \: - \bigg(0 - \frac{1}{2k + 1} \bigg) + \displaystyle\int_{0}^{\dfrac{\pi}{2}}\rm \frac{ {sin}^{2} kx}{sinx}dx$

$\rm \: = \: \bigg(\frac{1}{2k + 1} \bigg) + \displaystyle\int_{0}^{\dfrac{\pi}{2}}\rm \frac{ {sin}^{2} kx}{sinx}dx$

[ Because of substituting the value from Step 2 ]

$\rm \: = \: 1 + \dfrac{1}{3} + \dfrac{1}{5} + - - - + \dfrac{1}{2k + 1}$

$\rm\implies \:P(n) \: is \: true \: for \: n \: = \: k \: + \: 1$

Hence, By the Principle of Mathematical Induction, we get

$\rm \: \displaystyle\int_{0}^{ \dfrac{\pi}{2} }\rm \frac{ {sin}^{2} nx}{sinx} = 1 + \frac{1}{3} + \frac{1}{5} + - - + \frac{1}{2n - 1}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

FORMULA USED

$\displaystyle\int\rm sinx \: dx \: = \: - \: cosx \: + \: c \:$

$\rm \: cos(2n + 1)\dfrac{\pi}{2} = 0 \: where \: n \: is \: integer$