Mister Exam

Other calculators

Integral of cos^2t*sin^2t dt

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src]
  1                   
  /                   
 |                    
 |     2       2      
 |  cos (t)*sin (t) dt
 |                    
/                     
0                     
01sin2(t)cos2(t)dt\int\limits_{0}^{1} \sin^{2}{\left(t \right)} \cos^{2}{\left(t \right)}\, dt
Integral(cos(t)^2*sin(t)^2, (t, 0, 1))
Detail solution
  1. Rewrite the integrand:

    sin2(t)cos2(t)=(12cos(2t)2)(cos(2t)2+12)\sin^{2}{\left(t \right)} \cos^{2}{\left(t \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right) \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right)

  2. There are multiple ways to do this integral.

    Method #1

    1. Let u=2tu = 2 t.

      Then let du=2dtdu = 2 dt and substitute dudu:

      (18cos2(u)8)du\int \left(\frac{1}{8} - \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du

      1. Integrate term-by-term:

        1. The integral of a constant is the constant times the variable of integration:

          18du=u8\int \frac{1}{8}\, du = \frac{u}{8}

        1. The integral of a constant times a function is the constant times the integral of the function:

          (cos2(u)8)du=cos2(u)du8\int \left(- \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{8}

          1. Rewrite the integrand:

            cos2(u)=cos(2u)2+12\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}

          2. Integrate term-by-term:

            1. The integral of a constant times a function is the constant times the integral of the function:

              cos(2u)2du=cos(2u)du2\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}

              1. Let u=2uu = 2 u.

                Then let du=2dudu = 2 du and substitute du2\frac{du}{2}:

                cos(u)2du\int \frac{\cos{\left(u \right)}}{2}\, du

                1. The integral of a constant times a function is the constant times the integral of the function:

                  cos(u)du=cos(u)du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

                  1. The integral of cosine is sine:

                    cos(u)du=sin(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

                  So, the result is: sin(u)2\frac{\sin{\left(u \right)}}{2}

                Now substitute uu back in:

                sin(2u)2\frac{\sin{\left(2 u \right)}}{2}

              So, the result is: sin(2u)4\frac{\sin{\left(2 u \right)}}{4}

            1. The integral of a constant is the constant times the variable of integration:

              12du=u2\int \frac{1}{2}\, du = \frac{u}{2}

            The result is: u2+sin(2u)4\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}

          So, the result is: u16sin(2u)32- \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}

        The result is: u16sin(2u)32\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}

      Now substitute uu back in:

      t8sin(4t)32\frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}

    Method #2

    1. Rewrite the integrand:

      (12cos(2t)2)(cos(2t)2+12)=14cos2(2t)4\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right) \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right) = \frac{1}{4} - \frac{\cos^{2}{\left(2 t \right)}}{4}

    2. Integrate term-by-term:

      1. The integral of a constant is the constant times the variable of integration:

        14dt=t4\int \frac{1}{4}\, dt = \frac{t}{4}

      1. The integral of a constant times a function is the constant times the integral of the function:

        (cos2(2t)4)dt=cos2(2t)dt4\int \left(- \frac{\cos^{2}{\left(2 t \right)}}{4}\right)\, dt = - \frac{\int \cos^{2}{\left(2 t \right)}\, dt}{4}

        1. Rewrite the integrand:

          cos2(2t)=cos(4t)2+12\cos^{2}{\left(2 t \right)} = \frac{\cos{\left(4 t \right)}}{2} + \frac{1}{2}

        2. Integrate term-by-term:

          1. The integral of a constant times a function is the constant times the integral of the function:

            cos(4t)2dt=cos(4t)dt2\int \frac{\cos{\left(4 t \right)}}{2}\, dt = \frac{\int \cos{\left(4 t \right)}\, dt}{2}

            1. Let u=4tu = 4 t.

              Then let du=4dtdu = 4 dt and substitute du4\frac{du}{4}:

              cos(u)4du\int \frac{\cos{\left(u \right)}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                cos(u)du=cos(u)du4\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                1. The integral of cosine is sine:

                  cos(u)du=sin(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

                So, the result is: sin(u)4\frac{\sin{\left(u \right)}}{4}

              Now substitute uu back in:

              sin(4t)4\frac{\sin{\left(4 t \right)}}{4}

            So, the result is: sin(4t)8\frac{\sin{\left(4 t \right)}}{8}

          1. The integral of a constant is the constant times the variable of integration:

            12dt=t2\int \frac{1}{2}\, dt = \frac{t}{2}

          The result is: t2+sin(4t)8\frac{t}{2} + \frac{\sin{\left(4 t \right)}}{8}

        So, the result is: t8sin(4t)32- \frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}

      The result is: t8sin(4t)32\frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}

    Method #3

    1. Rewrite the integrand:

      (12cos(2t)2)(cos(2t)2+12)=14cos2(2t)4\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right) \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right) = \frac{1}{4} - \frac{\cos^{2}{\left(2 t \right)}}{4}

    2. Integrate term-by-term:

      1. The integral of a constant is the constant times the variable of integration:

        14dt=t4\int \frac{1}{4}\, dt = \frac{t}{4}

      1. The integral of a constant times a function is the constant times the integral of the function:

        (cos2(2t)4)dt=cos2(2t)dt4\int \left(- \frac{\cos^{2}{\left(2 t \right)}}{4}\right)\, dt = - \frac{\int \cos^{2}{\left(2 t \right)}\, dt}{4}

        1. Rewrite the integrand:

          cos2(2t)=cos(4t)2+12\cos^{2}{\left(2 t \right)} = \frac{\cos{\left(4 t \right)}}{2} + \frac{1}{2}

        2. Integrate term-by-term:

          1. The integral of a constant times a function is the constant times the integral of the function:

            cos(4t)2dt=cos(4t)dt2\int \frac{\cos{\left(4 t \right)}}{2}\, dt = \frac{\int \cos{\left(4 t \right)}\, dt}{2}

            1. Let u=4tu = 4 t.

              Then let du=4dtdu = 4 dt and substitute du4\frac{du}{4}:

              cos(u)4du\int \frac{\cos{\left(u \right)}}{4}\, du

              1. The integral of a constant times a function is the constant times the integral of the function:

                cos(u)du=cos(u)du4\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                1. The integral of cosine is sine:

                  cos(u)du=sin(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

                So, the result is: sin(u)4\frac{\sin{\left(u \right)}}{4}

              Now substitute uu back in:

              sin(4t)4\frac{\sin{\left(4 t \right)}}{4}

            So, the result is: sin(4t)8\frac{\sin{\left(4 t \right)}}{8}

          1. The integral of a constant is the constant times the variable of integration:

            12dt=t2\int \frac{1}{2}\, dt = \frac{t}{2}

          The result is: t2+sin(4t)8\frac{t}{2} + \frac{\sin{\left(4 t \right)}}{8}

        So, the result is: t8sin(4t)32- \frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}

      The result is: t8sin(4t)32\frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}

  3. Add the constant of integration:

    t8sin(4t)32+constant\frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}+ \mathrm{constant}


The answer is:

t8sin(4t)32+constant\frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}+ \mathrm{constant}

The graph
0.001.000.100.200.300.400.500.600.700.800.900.000.50
The answer [src]
1   cos(2)*sin(2)
- - -------------
8         16     
sin(2)cos(2)16+18- \frac{\sin{\left(2 \right)} \cos{\left(2 \right)}}{16} + \frac{1}{8}
=
=
1   cos(2)*sin(2)
- - -------------
8         16     
sin(2)cos(2)16+18- \frac{\sin{\left(2 \right)} \cos{\left(2 \right)}}{16} + \frac{1}{8}
1/8 - cos(2)*sin(2)/16
Numerical answer [src]
0.148650077978373
0.148650077978373

    Use the examples entering the upper and lower limits of integration.