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eistein+newton+gallileo=_me
 3 years ago
find the integral of 1/root(x^24)
eistein+newton+gallileo=_me
 3 years ago
find the integral of 1/root(x^24)

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myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1315245910188:dw

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1let \[\sec(\theta)=\frac{x}{2}\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1\[2 \sec(\theta)=x => 2\sec(\theta)\tan(\theta) d \theta=dx\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}\frac{1}{\sqrt{(2\sec(\theta))^24}} 2\sec(\theta) \tan(\theta) d \theta \]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{2}{\sqrt{4}}\int\limits_{}^{}\frac{\sec(\theta) \tan(\theta)}{\sqrt{\sec^2(\theta)1}} d \theta \]

eistein+newton+gallileo=_me
 3 years ago
Best ResponseYou've already chosen the best response.0mmmmm the answer is ln[x + root(X^2 4) ] + c

eistein+newton+gallileo=_me
 3 years ago
Best ResponseYou've already chosen the best response.0mmmmm the answer is ln[x + root(X^2 4) ] + c .......negative 4

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}\frac{\sec(\theta) \tan(\theta)}{\sqrt{\tan^2(\theta)}} d \theta\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}\sec(\theta) d \theta =\int\limits_{}^{}\sec(\theta) \cdot \frac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)} d \theta\] let \[u=\sec(\theta)+\tan(\theta)=>du=[\sec(\theta)\tan(\theta)+\sec^2(\theta)] d \theta\] so we have \[\int\limits_{}^{}\frac{du}{u}=\lnu+C\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1but \[u=\sec(\theta)+\tan(\theta)\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1so we have \[\ln\sec(\theta)+\tan(\theta)+C\] but we made the substition that \[\sec(\theta)=\frac{x}{2}\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1so if \[\sec(\theta)=\frac{x}{2}\] dw:1315246444736:dw so \[\tan(\theta)=\frac{\sqrt{x^24}}{2}\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1so we actually have \[\ln\frac{x}{2}+\frac{\sqrt{x^24}}{2}+C\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1\[\ln\frac{x+\sqrt{x^24}}{2}+C=\lnx+\sqrt{x^24}\ln2+C\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1but ln(2) is just a constant so it can all be included in the C part so we could write \[\lnx+\sqrt{x^24}+K\]
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