The integral ∫0π4cos2x+sin2x8xdx is equal to:
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Step-by-Step Solution
Step 1: Apply the property of definite integrals
We use the property of definite integrals which states that for an integral from 0 to a, we can replace x with a-x without changing the value of the integral. In this case, a=π. Let the given integral be I.
Step 2: Substitute x=π−x into the integral
Substitute x=π−x into the original integral. Recall that cos(π−x)=−cosx and sin(π−x)=sinx. Therefore, cos2(π−x)=(−cosx)2=cos2x and sin2(π−x)=(sinx)2=sin2x.
Step 3: Simplify the integral
After substituting and simplifying the trigonometric terms, the denominator remains unchanged. We can split this integral into two parts.
Step 4: Add the original and modified integrals
Adding the original integral I and the modified integral I (from the previous step) gives 2I. The denominators are the same, so we can combine the numerators.
Step 5: Combine and simplify the numerators
The 8x terms in the numerator cancel out, leaving 8π. Now we have a simpler integral to evaluate.
Step 6: Evaluate the simplified integral
Divide the numerator and denominator by cos2x. This transforms the integral into a form involving sec2x and tan2x, which can be solved using a substitution. Recall that sec2x=1+tan2x.
Step 7: Perform substitution and integrate
Let t=tanx. Then dt=sec2xdx. When x=0, t=tan0=0. When x=π, t=tanπ=0. However, the function tanx is discontinuous at x=π/2. We need to split the integral into two parts: from 0 to π/2 and from π/2 to π. Due to symmetry, ∫0π4+tan2xsec2xdx=2∫0π/24+tan2xsec2xdx. For the integral from 0 to π/2, when x=π/2, t→∞.
Step 8: Complete the integration
The integral ∫a2+t2dt=a1tan−1(at). Here a=2. So, ∫0π/24+tan2xsec2xdx=[21tan−1(2tanx)]0π/2=21tan−1(∞)−21tan−1(0)=21⋅2π−0=4π. Therefore, 2I=8π⋅2⋅4π=4π2.
Step 9: Solve for I
Divide by 2 to find the value of I.