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poizon [28]
3 years ago
11

HELP PLEASE

Mathematics
1 answer:
nasty-shy [4]3 years ago
5 0

Answer:

Part A: -2.5

Part B: -1

Part C: 1

Part D: 2.5

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4 years ago
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\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}

A standard approach would be the tangent half-angle substitution:

t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx

Then

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from which we get

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So the integral becomes

\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}

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and expand the integrand into its partial fractions:

\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)

We have

\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C

=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C

=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C

6 0
3 years ago
Determine if the functions are even, odd or neither.
marusya05 [52]
I'm pretty sure its neither just substitute the x for -2 and solve. There has to be a consistency in the values 3 times for there to be a pattern.
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3 years ago
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