Jee Mains Problem 1

Calculus Level 3

f ( x ) = s i n x + ∫ 0 x f ′ ( t ) ( 2 s i n t − s i n 2 t ) d t f(x) = sinx + \int_0^x f ' (t)(2sint - sin^{2}t)dt

then f(x) is

t a n x 1 − s i n x \frac{tanx}{1-sinx} c o s x 1 − c o s x \frac{cosx}{1-cosx} s i n x 1 − s i n x \frac{sinx}{1- sinx} x 1 − s i n x \frac{x}{1- sinx}

U Z by U Z , 24, Guyana

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Pranjal Jain
Dec 21, 2014

f ( x ) = sin ⁡ x + ∫ 0 x f ′ ( t ) ( 2 sin ⁡ t − sin ⁡ 2 t ) d t f(x)=\sin x+\displaystyle\int_{0}^{x} f'(t)(2\sin t-\sin^2 t) dt

Using Leibniz's formula,

f ′ ( x ) = cos ⁡ x + f ′ ( x ) ( 2 sin ⁡ x − sin ⁡ 2 x ) f'(x)=\cos x+f'(x)(2\sin x-\sin^2 x)

⇒ f ′ ( x ) = cos ⁡ x ( 1 − sin ⁡ x ) 2 \Rightarrow f'(x)=\dfrac{\cos x}{(1-\sin x)^{2}}

Substitute t = sin ⁡ x t=\sin x , g ( t ) = ∫ d t ( 1 − t ) 2 g(t)=\displaystyle\int \frac{dt}{(1-t)^{2}}

Integrate and substitute t = sin ⁡ x t=\sin x to get f ( x ) = sin ⁡ x 1 − sin ⁡ x \boxed{f(x)=\dfrac{\sin x}{1-\sin x}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I think it's also worth mentioning that setting x equal to zero in the given equation at the start allows us to identify f(0)=0 as the initial condition, which is necessary to find the final answer, instead of a general solution with an arbitrary constant.

Tristan Goodman - 2 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...