# Calculus

Use the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm confused about how to solve this one.

1. Let V = ∫sin^2x dx
Since cos^2x is a mirror image of sin^2x, ∫cos^2x dx = V

Now, since sin^2x+cos^2x = 1,

2V = ∫[0,π/2] 1 dx = π/2
V = π/4

check:

sin^2x = (1-cos2x)/2
∫sin^2x dx = ∫(1-cos2x)/2 dx
= 1/2 (x - 1/2 sin2x) [0,π/2]
= 1/2[(π/2)-(0)]
= π/4

posted by Steve

## Similar Questions

1. ### Calculus

Use the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm
2. ### TRIG!

Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x +
3. ### tigonometry

expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
4. ### Integral

That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten
5. ### Mathematics - Trigonometric Identities

Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) =
6. ### algebra

Can someone please help me do this problem? That would be great! Simplify the expression: sin theta + cos theta * cot theta I'll use A for theta. Cot A = sin A / cos A Therefore: sin A + (cos A * sin A / cos A) = sin A + sin A = 2
7. ### trig

it says to verify the following identity, working only on one side: cotx+tanx=cscx*secx Work the left side. cot x + tan x = cos x/sin x + sin x/cos x = (cos^2 x +sin^2x)/(sin x cos x) = 1/(sin x cos x) = 1/sin x * 1/cos x You're
8. ### Integration by Parts

integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be -pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=-cos(t) i integral sin(t)e^(it)dt= -e^(it)cos(t)+i*integral
9. ### trig

Reduce the following to the sine or cosine of one angle: (i) sin145*cos75 - cos145*sin75 (ii) cos35*cos15 - sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b) - sin(a)sin)(b) (1)The
10. ### trig integration

s- integral endpoints are 0 and pi/2 i need to find the integral of sin^2 (2x) dx. i know that the answer is pi/4, but im not sure how to get to it. i know: s sin^2(2x)dx= 1/2 [1-cos (4x)] dx, but then i'm confused. The indefinite

More Similar Questions