2017年2月17日 · Note that #16=2^4#. #rArr2^(x+2)=(2^4)^x# #rArrcolor(red)(2)^(x+2)=color(red)(2)^(4x)# Since the bases are equal, that is #color(red)(2)# …

2018年3月23日 · Perform the substitution. #x=4sintheta#, #=>#, #dx=4costhetad theta# #sqrt(16-x^2)=sqrt(16-16sin^2theta)=4costheta#

2018年6月1日 · Substitute #x=4sin theta# to turn the denominator into a simple trig function via the identity #sin^2 theta +cos^2 theta=1#. Note that #(dx)/(d theta)=4 cos theta# and that the …

c = 16 Substitute these values into the formula above to find that (0, 16) is the vertex point. The expression 16 - x^2 can be factored as a difference of squares. (4 + x)*(4 - x)=0 So, the x …

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2016年9月10日 · -sqrt(16-x^2)+C Although this is well set up for a shorter, non-trigonometric substitution (see: u=16-x^2), we can make the trigonometric substitution x=4sintheta. Note that …

2018年3月11日 · I find that problems like this are solved easier by applying our rules of logarithmic functions. #y = x*sqrt(16-x^2)# We know from the rules of algebra that manipulated …

The critical numbers are +-2sqrt2, and +-4. Critical numbers for a function, f, are numbers in the domain f at which f'(x) = 0 or f'(x) does not exist. For, f(x) = xsqrt(16 - x^2), the derivative is: …

2018年5月5日 · Here, #I=intx^3sqrt(16-x^2)dx# Let, #x=4sinu=>dx=4cosudu# #and sinu=x/4=>cosu=sqrt(1-sin^2u)=sqrt(1-x^2/16# ...

2018年1月28日 · Try #x=4 tan theta#.Then #x = 4 sec^2theta d theta# So #int x^3/sqrt{x^2+16} dx = int {64 tan^3 theta times 4 sec^2theta}/{4sectheta}d theta #