"Evaluate the integrals ∫0π4[sec⁡ti+tan2tj−tsin⁡tk]dt" was answered by our community

babeeb0oL

Answered question

2021-10-24

Evaluate the integrals

\int_{0}^{\frac{π}{4}} [\sec t i + \tan^{2} t j - t \sin t k] d t

Khribechy

Skilled2021-10-25Added 100 answers

To find the integral

\int_{0}^{\frac{π}{4}} [\sec t i + \tan^{2} t j - t \sin t k] d t

Explanation:

\int_{0}^{\frac{π}{4}} [\sec t i + \tan^{2} t j - t \sin t k] d t = [\int_{0}^{\frac{π}{4}} \sec t d t] i + [\int_{0}^{\frac{π}{4}} \tan^{2} t d t] j - [\int_{0}^{\frac{π}{4}} t \sin t d t] k

= I_{1} i + I_{2} j - I_{3} k

Where

I_{1} = \int_{0}^{\frac{π}{4}} \sec t d t, I_{2} = \int_{0}^{\frac{π}{4}}, I = \int_{0}^{\frac{π}{4}} \sin t d t

I_{1} = \int_{0}^{\frac{π}{4}} \sec t d t

On integrating

= {[\ln | \sec t + \tan t |]}_{0}^{\frac{π}{4}}

= [\ln | \sec (\frac{π}{4}) + \tan (\frac{π}{4}) | - \ln | \sec (0) + \tan (0) |]

= [\ln | \sqrt{2} + 1 | - \ln | 1 |]

= [\ln | \sqrt{2} + 1 |]

I_{2} = \int_{0}^{\frac{π}{4}} \tan^{2} t d t

= \int_{0}^{\frac{π}{4}} (\sec^{2} t - 1) d t

=

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