If int sqrt{1 + sin x} , dx = -4 cos(ax + b) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2 \sin(x/2) \cos(x/2) = (\sin(x/2) + \cos(x/2))^2$.Thus, $\sqrt{1 + \sin x} = |\sin(x/2) +

Vedclass · Accepted Answer

$1/2, \pi/4$

Vedclass · Answer

(B) We know that $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2 \sin(x/2) \cos(x/2) = (\sin(x/2) + \cos(x/2))^2$.Thus, $\sqrt{1 + \sin x} = |\sin(x/2) + \cos(x/2)|$.Assuming the interval where $\sin(x/2) + \cos(x/2) > 0$, we have $\int (\sin(x/2) + \cos(x/2)) \, dx$.$= -2 \cos(x/2) + 2 \sin(x/2) = 2(\sin(x/2) - \cos(x/2))$.Using $\sin 	heta - \cos 	heta = \sqrt{2} \sin(	heta - \pi/4) = -\sqrt{2} \cos(	heta + \pi/4)$.So, $2(\sin(x/2) - \cos(x/2)) = -2\sqrt{2} \cos(x/2 + \pi/4)$.Comparing this with $-4 \cos(ax + b)$, we note the coefficient mismatch. Let's re-evaluate: $\sqrt{1 + \sin x} = \sqrt{(\cos(x/2) + \sin(x/2))^2} = \cos(x/2) + \sin(x/2)$.Integral $= 2 \sin(x/2) - 2 \cos(x/2) = 2\sqrt{2} (\frac{1}{\sqrt{2}} \sin(x/2) - \frac{1}{\sqrt{2}} \cos(x/2)) = 2\sqrt{2} \sin(x/2 - \pi/4) = -2\sqrt{2} \cos(x/2 + \pi/4)$.Given the form $-4 \cos(ax+b)$, there is a constant factor discrepancy in the problem statement, but $a=1/2$ and $b=\pi/4$ match the argument of the cosine function.

If $\int \sqrt{1 + \sin x} \, dx = -4 \cos(ax + b) + c$, then the values of $a$ and $b$ respectively are:

Similar Questions

$\int \frac{x}{x + 2} dx$ is equal to

If $\int \frac{dx}{\sqrt{16-9x^2}} = A \sin^{-1}(Bx) + C$,then $A+B=$

Which one of the following is $TRUE$?

If $\int \frac{1}{\sqrt{9-16 x^2}} d x=\alpha \sin ^{-1}(\beta x)+c$,then $\alpha+\frac{1}{\beta}=$

Integrate the function $\sin (ax+b) \cos (ax+b)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module