"Evaluate the integral. ∫tdt7−t2" was answered by our experts

Khadija Wells

Answered question

2021-09-23

Evaluate the integral.

\int \frac{t d t}{\sqrt{7 - t^{2}}}

Layton

Skilled2021-09-24Added 89 answers

Step 1
Given:

I = \int \frac{t d t}{\sqrt{7 - t^{2}}}

for evaluating given integral, in given integral we substitute

\sqrt{7 - t^{2}} = x

...(1)
now, differentiating equation(1) with respect to x

\frac{d}{d x} (\sqrt{7 - t^{2}}) = \frac{d x}{d t} (∵ \frac{d}{d x} (\sqrt{x}) = \frac{1}{2 \sqrt{x}})

\frac{1}{2 \sqrt{7 - t^{2}}} \frac{d}{d t} (7 - t^{2}) = \frac{d x}{d t} (∵ \frac{d}{d x} (x^{n}) = n x^{n - 1})

\frac{1}{2 \sqrt{7 - t^{2}}} (0 - 2 t) = \frac{d x}{d t}

- \frac{2 t}{2 \sqrt{7 - t^{2}}} = \frac{d x}{d t}

\frac{t}{\sqrt{7 - t^{2}}} d t = - d x

Step 2
now, replacing

\frac{t}{\sqrt{7 - t^{2}}} d t

with (−dx) in given integral
so,

\int \frac{t d t}{\sqrt{7 - t^{2}}} = - \int d x

= - x + c (∵ x = \sqrt{7 - t^{2}})

= - \sqrt{7 - t^{2}} + c

hence, given integral is equal to

(- \sqrt{7 - t^{2}} + c)

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