If I=∫02πsin2x+cos2xsin21xdx, then ∫02Isin2x+cos4xxsinxcosxdx equals:
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Step-by-Step Solution
Step 1: Simplify the first integral
The denominator of the first integral is sin2x+cos2x. This is a fundamental trigonometric identity, which simplifies to 1. So, the integral I becomes ∫02πsin21xdx.
Step 2: Evaluate the first integral
To evaluate the integral, we find the antiderivative of sin21x, which is −2cos21x. Then, we apply the limits of integration from 0 to 2π.
Step 3: Calculate the value of I
Substitute the values of cos4π=21 and cos0=1 into the expression. Simplify the result to find the value of I.
Step 4: Define the second integral and apply property
Let the second integral be J. We substitute the value of I we just calculated into the upper limit of the integral. The integral is of the form ∫0af(x)dx.
Step 5: Apply the property ∫0af(x)dx=∫0af(a−x)dx
We use the property of definite integrals that ∫0af(x)dx=∫0af(a−x)dx. Here, a=2I. This property is often useful when the integrand has symmetry.
Step 6: Observe the symmetry of the integrand
The value of 2I=2(2−2)≈1.172. This is not a standard angle like π or 2π. For the property sin(a−x)=sinx and cos(a−x)=−cosx to hold, a must be π. If we assume the problem intended 2I=π, then sin(π−x)=sinx and cos(π−x)=−cosx. Substituting these into the integral, we get a new expression for J.
Step 7: Add the original and modified integrals (assuming 2I=π)
Adding the original integral J and the modified integral J (after applying the property), we combine the numerators. This step is part of a common technique to solve integrals of this form, but it requires the upper limit to be π for the trigonometric identities to simplify the expression further.