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\[ \int \dfrac{\sin x + \sin^3 x}{\cos (2x) } \, dx = \,? \]

Note by Brilliant Member 1 year, 11 months ago

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Hint: Start with \( y = \cos x\), then do long division, and you get a form of \( \int \frac 1{z^2-a^2} \, dz \).

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thank you very much big brother, i am able to do it now. the answer coming is " right"

Pi Han Goh is correct , Let \( t=cosx \implies dx=\dfrac{dt}{-\sin{x}}\)

\(I=\displaystyle\int \dfrac{t^2-2}{2t^2-1}\cdot dt\)

yeah i totally agree with you and pi han goh . thanks for seeing to it, plz keep supporting.

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestHint: Start with \( y = \cos x\), then do long division, and you get a form of \( \int \frac 1{z^2-a^2} \, dz \).Log in to reply

thank you very much big brother, i am able to do it now. the answer coming is " right"

Log in to reply

Pi Han Goh is correct , Let \( t=cosx \implies dx=\dfrac{dt}{-\sin{x}}\)

\(I=\displaystyle\int \dfrac{t^2-2}{2t^2-1}\cdot dt\)

Log in to reply

yeah i totally agree with you and pi han goh . thanks for seeing to it, plz keep supporting.

Log in to reply