Trig Integral Example

[Jason76]

\(\displaystyle \int \sec^{2}(\dfrac{t}{6}) dt\)


\(\displaystyle \int \sec^{2}(u) dt\)

\(\displaystyle u = \dfrac{t}{6}\)

\(\displaystyle du = ? \)

Is \(\displaystyle du\) this?

\(\displaystyle \dfrac{[(6)(1) - (t)(0)]}{[6]^{2}} = \dfrac{6}{36} = \dfrac{1}{6}\)

\(\displaystyle du = \dfrac{1}{6}\)

\(\displaystyle 6 du = dt\)

\(\displaystyle 6 \int (u) dt\)

\(\displaystyle 6 \tan(\dfrac{t}{6}) + C\)

 

[lookagain]

\(\displaystyle \int \sec^{2}(\dfrac{t}{6}) dt\)


\(\displaystyle \int \sec^{2}(u) dt \ \ \ \ \ \)Don't mix u and dt together.

\(\displaystyle u = \dfrac{t}{6} \ \ \ \ \ \ \) This line should come *before* any substitutions.

\(\displaystyle du = ? \ \ \ \ \)Don't solve for du. Solve for t. Then find dt.

Is \(\displaystyle du\) this?

\(\displaystyle \dfrac{[(6)(1) - (t)(0)]}{[6]^{2}} = \dfrac{6}{36} = \dfrac{1}{6} \ \ \ \ \ \) No. Don't do this.

\(\displaystyle du = \dfrac{1}{6} \ \ \ \ \ \) No.

\(\displaystyle 6 du = dt \ \ \ \ \ \)Yes, but this does not follow from your line immediately above.

\(\displaystyle 6 \int (u) dt \ \ \ \ \ \ \)No. Again, don't mix u and dt. You are missing the sec^2 part. What else is messed up?

\(\displaystyle 6 \tan(\dfrac{t}{6}) + C\) \(\displaystyle \ \ \ \ \ \)The work doesn't support this answer.

Fudging or the appearance of fudging is present.

\(\displaystyle \int \sec^{2}(\dfrac{t}{6}) dt\)

\(\displaystyle Let \ \ u \ = \ \dfrac{t}{6} \)

\(\displaystyle Then \ \ t \ = \ 6u \)

\(\displaystyle dt \ = \ 6 \ du\)

\(\displaystyle \int [\sec^{2}(u)] 6 \ du \)


\(\displaystyle 6\int \sec^{2}(u) du \)

\(\displaystyle 6 \ tan(u) \ + \ C\)

\(\displaystyle 6 \ tan\bigg(\dfrac{t}{6}\bigg) \ + \ C \)