Let g(x) be a differentiable function satisfying ddx{g(x)}=g(x) and g(0)=1, then ∫g(x)2−sin⁡2x1−cos⁡2xdx is equal to ag(x)cot(bx)+c

g′(x)g(x)=1⇒ln⁡(g(x))=x+c⇒g(x)=excg(0)=1⇒c=1 ∫g(x)2−sin⁡2x1−cos⁡2xdx =∫ex2−2sin⁡xcosx2sin2xdx =∫excosec2x-cotxdx =ex-cotx+csince∫exf(x)+f′(x)dx=exf(x)+ctherefore a=-1,b=1

Let $$g(x)$$ be a differentiable function satisfying $$ddx{g(x)}=g(x)$$ and $$g(0)=1$$, then ∫$$g(x)2$$−sin⁡2x1−cos⁡2xdx is equal to $$ag(x)cot(bx)+c$$

g′(x)g(x)=1⇒ln⁡(g(x))=x+c⇒g(x)=excg(0)=1⇒c=1 ∫g(x)2−sin⁡2x1−cos⁡2xdx =∫ex2−2sin⁡xcosx2sin2xdx =∫excosec2x-cotxdx =ex-cotx+csince∫exf(x)+f′(x)dx=exf(x)+ctherefore a=-1,b=1

Questions

Let g(x) be a differentiable function satisfying $\frac{d}{d x} {g (x)} = g (x) and g (0) = 1, then \int g (x) (\frac{2 - \sin 2 x}{1 - \cos 2 x}) d x$ is equal to ag(x)cot(bx)+c

Remember concepts with our Masterclasses.

80k Users

60 mins Expert Faculty Ask Questions

a+b=0

a^{2} + b^{2} = 2

a^{2} + 2 b = 3

2 a^{2} - b^{2} = - 1

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

detailed solution

Correct option is A

$\begin{array}{l} \frac{g^{'} (x)}{g (x)} = 1 \Rightarrow \ln (g (x)) = x + c \Rightarrow g (x) = e^{x} c \\ g (0) = 1 \Rightarrow c = 1 \int g (x) (\frac{2 - \sin 2 x}{1 - \cos 2 x}) d x = \int e^{x} (\frac{2 - 2 \sin x \cos x}{2 \sin^{2} x}) d x = \int e^{x} (c o s e c^{2} x - c o t x) d x = e^{x} (- c o t x) + c \\ (\sin c e \int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + c) \end{array}$
therefore a=-1,b=1