What is #f(x) = int sin^2x-cotx dx# if #f((5pi)/4) = 0 #?

Answer 1

#f(x)=x/2-(sin2x)/4-ln|sinx|+1/4-5pi/8-1/2ln2.#

#f(x)=int(sin^2x-cotx)dx.#
#:. f(x)=intsin^2xdx-intcotxdx,#
#=int(1-cos2x)/2dx-ln|sinx|,#
#=1/2{int1dx-intcos2xdx}-ln|sinx|,#
#rArr f(x)=x/2-(sin2x)/4-ln|sinx|+C.#
#"To determine "C," we use the given cond. that "f(5pi/4)=0.#
#rArr 5pi/8-(sin(5pi/2))/4-ln|sin(5pi/4)|+C=0.#
#rArr 5pi/8-1/4-ln|-1/sqrt2|+C=0.#
#:. C=1/4-5pi/8-1/2ln2.#
Hence, #f(x)=x/2-(sin2x)/4-ln|sinx|+1/4-5pi/8-1/2ln2.#

Enjoy Maths.!

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To solve the integral ( f(x) = \int \sin^2(x) - \cot(x) , dx ) given that ( f\left(\frac{5\pi}{4}\right) = 0 ), we need to evaluate the integral and then find the constant of integration using the given condition.

Integrating ( \sin^2(x) - \cot(x) ) with respect to ( x ) yields: [ \int \sin^2(x) - \cot(x) , dx = \frac{x}{2} + \frac{\cos(2x)}{4} - \ln|\sin(x)| + C ]

Now, using the condition ( f\left(\frac{5\pi}{4}\right) = 0 ), we substitute ( x = \frac{5\pi}{4} ) into the expression for ( f(x) ) and solve for the constant of integration ( C ):

[ 0 = \frac{\frac{5\pi}{4}}{2} + \frac{\cos\left(\frac{5\pi}{2}\right)}{4} - \ln|\sin\left(\frac{5\pi}{4}\right)| + C ]

[ 0 = \frac{5\pi}{8} + \frac{0}{4} - \ln\left|\frac{\sqrt{2}}{2}\right| + C ]

[ 0 = \frac{5\pi}{8} - \frac{\ln(\sqrt{2}/2)}{4} + C ]

[ C = \frac{\ln(\sqrt{2}/2)}{4} - \frac{5\pi}{8} ]

Thus, the function ( f(x) ) is: [ f(x) = \frac{x}{2} + \frac{\cos(2x)}{4} - \ln|\sin(x)| + \frac{\ln(\sqrt{2}/2)}{4} - \frac{5\pi}{8} ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To find ( f(x) = \int \sin^2(x) - \cot(x) , dx ) when ( f\left(\frac{5\pi}{4}\right) = 0 ), we first integrate the function:

[ \int \sin^2(x) - \cot(x) , dx ]

[ = \int \sin^2(x) - \frac{\cos(x)}{\sin(x)} , dx ]

[ = \int \sin^2(x) - \frac{1}{\tan(x)} , dx ]

[ = \int \sin^2(x) - \frac{\sin(x)}{\cos(x)} , dx ]

[ = \int \sin^2(x) - \tan(x) , dx ]

Now, integrate term by term.

[ = \left( -\frac{\cos(2x)}{4} + \ln|\sin(x)| \right) + C ]

Given that ( f\left(\frac{5\pi}{4}\right) = 0 ), we substitute ( x = \frac{5\pi}{4} ) into the integrated function and set it equal to 0:

[ 0 = -\frac{\cos\left(\frac{5\pi}{2}\right)}{4} + \ln|\sin\left(\frac{5\pi}{4}\right)| + C ]

[ 0 = -\frac{0}{4} + \ln\left|\frac{\sqrt{2}}{2}\right| + C ]

[ 0 = 0 + \ln\left(\frac{\sqrt{2}}{2}\right) + C ]

[ C = -\ln\left(\frac{\sqrt{2}}{2}\right) ]

So, the function ( f(x) = \int \sin^2(x) - \cot(x) , dx ) with the condition ( f\left(\frac{5\pi}{4}\right) = 0 ) is:

[ f(x) = -\frac{\cos(2x)}{4} + \ln|\sin(x)| - \ln\left(\frac{\sqrt{2}}{2}\right) ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7