Find M in \int_{-\frac{\pi}{2}}^{M} \cos x dx= 1.5 with Expert Tips

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To solve the integral \(\int_{-\frac{\pi}{2}}^{M} \cos x \, dx = 1.5\), the anti-derivative of \(\cos x\) is \(\sin x\). Evaluating this from \(-\frac{\pi}{2}\) to \(M\) gives \(\sin M - (-1) = 1.5\). This simplifies to \(\sin M + 1 = 1.5\), leading to \(\sin M = 0.5\). The solution for \(M\) can be found using the inverse sine function, confirming the calculations.
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Hey what's up..I need a little help with the following problem.

If \int_{-\frac{\pi}{2}}^{M} \cos x dx= 1.5}, find M.

How do I go about doing this? I know that...the anti derivative of cos is sin, and that \sin x \right]_{-\frac{pi}{2}}^{M} = 1.5

Any help would be greatly appreciated...
 
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\sin x |^M_{-\pi/2} = \sin M - (-\pi/2) = 1.5

Are you ok from there?
 
ahhh..DUH! goshh...how did i NOT see that? Thanks man...haha...
 
It was a typo/error in post #2.

\sin x\left |_{-\frac{\pi}{2}}^{M}\right =1.5\Rightarrow \sin M+\sin\frac{\pi}{2}=1.5\Rightarrow \sin M +1 =1.5

Daniel.
 
Oops! You're right, of course.
 
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