积分∫₀^(π/4) (cos2x)^(3/2)cosx dx计算过程的错误排查咨询

Hey there! It's great that you landed on the correct answer using the substitution ( t = \frac{\sin\theta}{\sqrt{2}} ), but figuring out where your initial method went wrong is totally doable—we just need a bit more info first. Could you share the step-by-step work from the method you used originally? That way we can pinpoint exactly where things might have slipped up.

In the meantime, here are some common pitfalls that often trip people up when working on integrals like this:

• Incorrect substitution domain: When you swap out variables, make sure the new limits of integration match the original interval. For example, if you used a substitution like ( u = \cos2x ), double-check that you adjusted the bounds correctly from ( x=0 ) to ( x=\pi/4 ).
• Mistakes in trigonometric identities: ( \cos2x ) has multiple forms (( 1-2\sin^2x ), ( 2\cos^2x-1 ), ( \cos^2x-\sin2x ))—if you picked the wrong one or messed up expanding it with ( \cos x ), that could throw off the entire integral.
• Sign errors with radicals: Since ( (\cos2x)^{3/2} = (\sqrt{\cos2x})^3 ), you have to ensure that ( \cos2x ) is non-negative over the interval (which it is here, since ( 2x ) goes from 0 to ( \pi/2 )), but if you manipulated the radical incorrectly (like pulling out terms without considering sign), that might cause issues.
• Algebraic slip-ups during integration: When expanding the integrand after substitution, it's easy to make a mistake in multiplying terms or combining exponents. Even a small error here can lead to a wrong final result.
• Forgot to adjust for substitution differential: If you used a substitution like ( u = \sin x ), remember that ( du = \cos x dx )—but if you didn't account for how ( \cos2x ) translates to ( u ) properly, that's a common mistake.

Once you share your original steps, we can dive into the exact error together!

备注：内容来源于stack exchange，提问作者Daksh