x/sqrt(x)=sqrt(x) integral sqrt(x)=2/3x3/2
ln(sinx) + 1/3ln(sin3x) + C
I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)Taking the natural log of both sides:ln(y)ln(x) * ln(x)ln(y) ln(x)^2Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * yFinally, substitute in the first equation, y x^ln(x):y' 2 ln(x) / x * x^ln(x)y'2 ln(x) * x ^ (ln(x) - 1)Sorry if everything is formatted really badly, this is my first post on answers.com.
Trying to integrate: cos2x sin x dx Substitute y = cos x Then dy = -sin x dx So the integral becomes: -y2dy Integrating gives -1/3 y3 Substituting back: -1/3 cos3x
x^(ln(2)/ln(x)-1)
The integral of cot(x)dx is ln|sin(x)| + C
x-1 = 1/x ∫1/x dx = ln x + C
1/2(x-ln(sin(x)+cos(x)))
This is the same as 4x-1, so the answer is just 4 ln x + C.
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
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Do you mean ln(x-2), or ln(x)-2? If it is ln(x-2): 1/(x-2) If it is ln(x)-2: 1/x
x/sqrt(x)=sqrt(x) integral sqrt(x)=2/3x3/2
dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).
ln(sinx) + 1/3ln(sin3x) + C
- ln ((x^2)-4)
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