How do you integrate #x * cos^2 (x)#?

Answer 1

#x^2/4+(x/4)sin2x+cos(2x)/8+c#.where c is integration constant .

#int (xcos^2x)dx# #=(1/2)int(x2cos^2x)dx# now #int(x2cos^2x)dx# =#int{x(1+cos2x)}dx# =#intxdx+int(xcos2x)dx# [integration by parts] =#x^2/2+x(int(cos2x)dx)-int[(dx/dx)int(cos2x)dx]dx# =#x^2/2+(xsin(2x))/2##-int((sin2x)/2)dx# =#x^2/2+(xsin(2x))/2##+(cos2x)/4##+c# hence the value of the integral :#x^2/4+(xsin(2x))/4+cos(2x)/8+c#,where c is constant of integration.

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Answer 2

Evelyn Agard

To integrate x * cos^2(x), you can use integration by parts. Let u = x and dv = cos^2(x) dx. Then differentiate u to get du/dx = 1, and integrate dv to get v = (1/2) * (x + sin(2x)/2). Apply the integration by parts formula:

∫x * cos^2(x) dx = u * v - ∫v * du

Substitute the values:

= x * (1/2) * (x + sin(2x)/2) - ∫(1/2) * (x + sin(2x)/2) * 1 dx

Then, integrate the remaining integral:

= (1/2) * (x^2/2 + x * sin(2x)/2) - (1/4) * (x^2/2 + x * sin(2x)/2 + cos(2x)/4) + C

Simplify to get the final result:

= (x^2/4) + (x * sin(2x)/4) - (cos(2x)/8) + C

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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.