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How do you use the substitution method? - Answers

When finding an antiderivative, substitution can be very helpful. In your integral expression, look for a term that, if you took the derivative and perhaps manipulated it a bit, would give you another term in the same equation. For example, if your integral is ∫tanxdx or ∫((sinx)/(cosx))dx , you can choose cosx as your arbitrary variable (often u or w). If u=cosx, du=-sinxdx. Therefore, your expression using u becomes -∫du/u. We know that the antiderivative of 1/u is ln (absolute value x) +C. So, our answer with u is -ln (abs. val. u) +C. But BE CAREFUL, this is not your final answer. You must plug in the value you originally designated as u. Therefore, the final answer is -ln (abs. val. cosx) +C, with C as an arbitrary constant.



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How do you use the substitution method? - Answers

https://math.answers.com/calculus/How_do_you_use_the_substitution_method

When finding an antiderivative, substitution can be very helpful. In your integral expression, look for a term that, if you took the derivative and perhaps manipulated it a bit, would give you another term in the same equation. For example, if your integral is ∫tanxdx or ∫((sinx)/(cosx))dx , you can choose cosx as your arbitrary variable (often u or w). If u=cosx, du=-sinxdx. Therefore, your expression using u becomes -∫du/u. We know that the antiderivative of 1/u is ln (absolute value x) +C. So, our answer with u is -ln (abs. val. u) +C. But BE CAREFUL, this is not your final answer. You must plug in the value you originally designated as u. Therefore, the final answer is -ln (abs. val. cosx) +C, with C as an arbitrary constant.



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https://math.answers.com/calculus/How_do_you_use_the_substitution_method

How do you use the substitution method? - Answers

When finding an antiderivative, substitution can be very helpful. In your integral expression, look for a term that, if you took the derivative and perhaps manipulated it a bit, would give you another term in the same equation. For example, if your integral is ∫tanxdx or ∫((sinx)/(cosx))dx , you can choose cosx as your arbitrary variable (often u or w). If u=cosx, du=-sinxdx. Therefore, your expression using u becomes -∫du/u. We know that the antiderivative of 1/u is ln (absolute value x) +C. So, our answer with u is -ln (abs. val. u) +C. But BE CAREFUL, this is not your final answer. You must plug in the value you originally designated as u. Therefore, the final answer is -ln (abs. val. cosx) +C, with C as an arbitrary constant.

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      When finding an antiderivative, substitution can be very helpful. In your integral expression, look for a term that, if you took the derivative and perhaps manipulated it a bit, would give you another term in the same equation. For example, if your integral is ∫tanxdx or ∫((sinx)/(cosx))dx , you can choose cosx as your arbitrary variable (often u or w). If u=cosx, du=-sinxdx. Therefore, your expression using u becomes -∫du/u. We know that the antiderivative of 1/u is ln (absolute value x) +C. So, our answer with u is -ln (abs. val. u) +C. But BE CAREFUL, this is not your final answer. You must plug in the value you originally designated as u. Therefore, the final answer is -ln (abs. val. cosx) +C, with C as an arbitrary constant.
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