*Notes to a video lecture on http://www.unizor.com*

__Indefinite Integral -__

Variable Substitution

Variable Substitution

Assume, you know that

**∫ f(x)**d**x = g(x) + C**What immediately follows from this is that the derivative of

*is the original function*

**g(x)***:*

**f(x)**

**g**^{I}(x) = f(x)or, equivalently,

*d*

**g(x) = f(x)**d**x**Given that, consider a derivative of a compound function

*:*

**g(w(x))****[**

*g(w(x))*]^{I}= f(w(x))·w^{I}(x)or, equivalently,

*d*

**g(w(x)) = f(w(x))·w**d^{I}(x)**x = f(w(x))**d**w(x)**From equality of derivatives or differentials of two functions

**f(w(x))**d**w(x) =**d**g(w(x))**follows that original functions (before the differentiation or obtained from the differential by integration) differ only by a constant.

Therefore,

**∫ f(w(x))**d**w(x) = ∫**d**g(w(x))**Since diiferentiation and integration are inversed to each other, the right part represents a function

*(plus constant, as usually with integration).*

**g(w(x))**Therefore,

**∫ f(w(x))**d**w(x) = g(w(x)) + C**or, equivalently,

**∫ f(w(x))·w**d^{I}(x)**x = g(w(x)) + C**Compare this with original relationship between

*and*

**f(x)***above:*

**g(x)**

**∫ f(x)**d**x = g(x) + C**As we see, to integrate

*, it is sufficient to integrate*

**f(w(x))·w**^{I}(x)*and substitute*

**f(x)***instead of*

**w(x)***in the answer.*

**x**This is a

**of integration.**

*substitution rule**Example 1*:

**∫ x·e**d^{(x²)}**x**To find this integral, notice that we can do the following:

**f(x) = e**^{x}

**∫ f(x)**d**x = e**^{x}+ C

**w(x) = x²**

**w**^{I}(x) = 2x

**f(w(x)) = e**^{(x²)}

**f(w(x))·w**^{I}(x) = e^{(x²)}·2x = e^{(x²)}·(x²)^{I}

**f(w(x))**d**w(x) = e**d^{(x²)}**(x²)**Using the above substitution, we obtain the answer:

**∫ x·e**d^{(x²)}**x = (1/2)∫ e**d^{(x²)}**(x²) = (1/2)e**^{(x²)}+ C*Checking by differentiating the answer (D*:

_{x}is the operation of differentiation)*D*

_{x}**(1/2)e**D^{(x²)}= (1/2)e^{(x²)}·_{x}**x² = (1/2)e**^{(x²)}·2x = x·e^{(x²)}The end

*Example 2*:

**∫ sin(x)·cos²(x)**d**x**To find this integral, notice that we can do the following:

**f(x) = x²**

**∫ f(x)**d**x = x³/3 + C**

**w(x) = cos(x)**

**w**^{I}(x) = −sin(x)

**f(w(x)) = cos²(x)**

**f(w(x))·w**^{I}(x) = −cos²(x)·sin(x) = cos²(x)·(cos(x))^{I}

**f(w(x))**d**w(x) = −cos²(x)**d**cos(x)**Using the above substitution, we obtain the answer:

**∫ sin(x)·cos²(x)**d**x = −∫ cos²(x)**d**cos(x) = −cos³(x)/3 + C***Checking by differentiating the answer*:

*D*

_{x}**−cos³(x)/3 = −3cos²(x)·(−sin(x))/3 = cos²(x)·sin(x)**The end.

*Example 3*:

**∫ ln(sin(x))·cos(x)**d**x**To find this integral, notice that

*,*

**(sin(x))**^{I}= cos(x)we use

*as an inner function:*

**sin(x)**

**ln(sin(x))·cos(x) = ln(sin(x))·(sin(x))**^{I}Also recall from the previous lecture that

**∫ ln(x)**d**x = x·ln(x) − x + C**Using the above substitution, we obtain the answer:

**∫ ln(sin(x))·cos(x)**d**x = ∫ ln(sin(x))**d**sin(x) = sin(x)·ln(sin(x)) − sin(x) + C***Checking by differentiating the answer*:

*D*

_{x}**(sin·ln(sin(x)) − sin(x)) = ln(sin(x))·cos(x) + sin(x)·(1/sin(x))·cos(x) − cos(x) = ln(sin(x))·cos(x)**The end.

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