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integral (0 to x) of sin(t)
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\[\int\limits_0^x\sin(t)\mathrm dt\]
How do I solve that?
if you find a standard table of integrals you will see that \[\large\boxed{\int\sin (ax)\mathrm dx=-\frac1a\cos (ax)}\]
so that means we can do this \[\int\limits_0^x\sin(t)\mathrm dt=-\cos(t)\big|_0^x\]
=-cos(x)+1
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Thanks
do you know the next step @mulline ?
\[f(x)\big|_a^b=f(b)-f(a)\]
so \[-\cos(t)\big|_0^x=\big(-\cos(x)\big)-\big(-\cos(0)\big)\]
@agostino direct answers are not helpful
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@UnkleRhaukus One can always check whether his/her answer is correct to see the validity of his/her knowledge.
Yes I understand how it is done I just making sure I was on the right track
mulline, you said you didn't know how to solve the question
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